{"id":8600,"date":"2019-11-26T12:57:02","date_gmt":"2019-11-26T07:27:02","guid":{"rendered":"https:\/\/kseebsolutions.guru\/?p=8600"},"modified":"2021-07-02T15:44:57","modified_gmt":"2021-07-02T10:14:57","slug":"1st-puc-chemistry-question-bank-chapter-1","status":"publish","type":"post","link":"https:\/\/kseebsolutions.guru\/1st-puc-chemistry-question-bank-chapter-1\/","title":{"rendered":"1st PUC Chemistry Question Bank Chapter 1 Some Basic Concepts of Chemistry"},"content":{"rendered":"

You can Download Chapter 1 Some Basic Concepts of Chemistry Questions and Answers, Notes, 1st PUC Chemistry Question Bank with Answers<\/a> Karnataka State Board Solutions help you to revise complete Syllabus and score more marks in your examinations.<\/p>\n

Karnataka 1st PUC Chemistry Question Bank Chapter 1 Some Basic Concepts of Chemistry<\/h2>\n

1st PUC Chemistry\u00a0Some Basic Concepts of Chemistry One Mark Questions and Answers<\/h3>\n

Question 1.
\nWhat is the unit for equivalent weight ?
\nAnswer:
\nEquivalent weight is a mere number, it has no units. When expressed in grams it is called gram equivalent weight.<\/p>\n

Question 2.
\nDefine Atomic weight.
\nAnswer:
\nThe atomic weight of an element is the ratio of average mass of an atom of an element to \\(\\frac{1}{12}\\) of mass of an atom of C-12 Isotope.<\/p>\n

Question 3.
\nWrite the mathematical form of Atomic weights.
\nAnswer:
\n\"1st
\nQuestion 4.
\nMention the unit for equivalent weight.
\nAnswer:
\nThere is no unit for atomic weight, when expressed in grams it is called gram atomic weight.<\/p>\n

\"KSEEB<\/p>\n

Question 5.
\nWrite the Relationship between Atomic weight and Valency weight.
\nAnswer:
\n\"1st<\/p>\n

Question 6.
\nDefine Valency.
\nAnswer:
\nValency : The combining capacity of an atom (or the number of electron present in the outer most orbital of an atom] which involves in bond formation.<\/p>\n

Question 7.
\nMention the methods to determine Equivalent weight.
\nAnswer:<\/p>\n

    \n
  • Oxide Method<\/li>\n
  • Chloride Method<\/li>\n
  • Hydrogen Displacement method<\/li>\n
  • Inter conversion method.<\/li>\n<\/ul>\n

    Question 8.
    \nDefine Molecular weight.
    \nAnswer:
    \nIt is defined as \u201c The ratio of mass of 1 molecule of the substance to \\(\\frac{1}{12}\\) mass of an atom of C-12 isotope”.<\/p>\n

    \"KSEEB<\/p>\n

    Question 9.
    \nWrite mathematical forms of Molecular weight.
    \nAnswer:
    \n\"1st<\/p>\n

    Question 10.
    \nDefine Avagadro\u2019s Law.
    \nAnswer:
    \nEqual volume of all gases and vapours under the same condition of temperature and pressure contains equal number of molecules.<\/p>\n

    Question 11.
    \nDefine Mole.
    \nAnswer:
    \n1 mole represent the amount of substance which contain Avagadro number of particles.<\/p>\n

    Question 12.
    \nWhat you mean by Avagadro Number ?
    \nAnswer:
    \nThe number of molecules present in 1 gm molecular weight of any substance.
    \nOR
    \nThe number of atoms present in 1 gm atomic weght of any element. It is represented by the symbol N – 6.022 \u00d7 1023<\/sup><\/p>\n

    Question 13.
    \nDefine Molality.
    \nAnswer:
    \nThe number of moles of solute dissolved in 1 kg of the solution.<\/p>\n

    \"KSEEB<\/p>\n

    Question 14.
    \nDefine Molarity.
    \nAnswer:
    \nNumber of molecular equivalent of solute dissolved in 1 dm3<\/sup> of the solution.<\/p>\n

    Question 15.
    \nWhat is mole fraction?
    \nAnswer:
    \nThe ratio of number of moles of a component to the number of moles present in the solution.<\/p>\n

    Question 16.
    \nWhat do you mean by Normal solution?
    \nAnswer:
    \nA solution containing 1 gm equivalent of the substance dissolved in 1 dm3<\/sup> of the solution.<\/p>\n

    Question 17.
    \nWhat you mean by decinormal solution?
    \nAnswer:
    \nA decinormal solution is the solution containing 0.1 g equivalent (\\(\\frac { 1 }{ 10 }\\) th<\/sup>) of solute \/ substance dissolved in 1 dm3<\/sup> of solution.<\/p>\n

    Question 18.
    \nWhat are indicators ?
    \nAnswer:
    \nIndicators are the chemical ubstance used to indicate the end point of a volumetric reaction by the change in colour.<\/p>\n

    Question 19.
    \nMention the types of Indicators.
    \nAnswer:<\/p>\n

      \n
    1. Internal indicator<\/li>\n
    2. External indicator<\/li>\n
    3. Self indicator<\/li>\n<\/ol>\n

      \"KSEEB<\/p>\n

      Question 20.
      \nDefine empirical formula.
      \nAnswer:
      \nIt is a formula which gives the relative number of atoms of each element in the compound.<\/p>\n

      Question 21.
      \nDefine Molecular formula.
      \nAnswer:
      \nIt is a formula which gives the actual number of atoms of each element present in the molecule of the given compound.<\/p>\n

      Question 22.
      \nWhat is the relation between empirical and molecular formula ?
      \nAnswer:
      \n\"1st<\/p>\n

      Question 23.
      \nState the law of definite proportion.
      \nAnswer:
      \nThe law states that “a chemical compound, however prepared, contains the same elements combined together in the same fixed proportion by mass”.<\/p>\n

      Question 24.
      \nDefine law of conservation of mass.
      \nAnswer:
      \nIt states that matter can neither be, created nor be destroyed.<\/p>\n

      Question 25.
      \nState Gay Lussac\u2019s law of combiningh volumes.
      \nAnswer:
      \nThe law sttes that \u2019when gaseous reactants combine to give gaseous products, they do so in volumes which bear a simple whole number ratio to one another and to the volume of producs, provided the volumes are measured under the similar conditions of temperature and pressure”.<\/p>\n

      Question 26.
      \nWhat is NTP ?
      \nAnswer:
      \n(Normal temperature and pressure or standard temperature and pressure). The temp 273K (0 \u00b0C) and pressure 101.3Kpa (or N\/m2<\/sup>) or latm = 760 mm of Hg level are taken as STP or NTP.<\/p>\n

      \"KSEEB<\/p>\n

      Question 27.
      \nWhat physical quantities are represented by the following units and what are their most common names?
      \n(i) kg ms-2<\/sup>
      \n(ii) kg m2<\/sup>s-2<\/sup>
      \n(iii) dm3<\/sup>
      \nAnswer:
      \n(i) Force (newton)
      \n(ii) Work (joule)
      \n(iii) Volume<\/p>\n

      Question 28.
      \nRewrite the following after required corrections : (i) The length of a rod is 10 cms, (ii) the work done by a system is 10 Joules.
      \nAnswer:
      \n(i) The length of a rod is 10 cm (s is not used) (ii) The work done by a system is 10 joules (small letter is used in place of capital letter).<\/p>\n

      \"KSEEB<\/p>\n

      Question 29.
      \nClassify the following substances into elements, compounds and mixtures:
      \n(i) Milk
      \n(ii) 22-carat gold
      \n(iii) Iodized table salt
      \n(iv) Diamond
      \n(v) Smoke
      \n(vi) Steel
      \n(vii) Brass
      \n(viii) Dry ice
      \n(ix) Mercury
      \n(x) Air
      \n(xi) Aerated drinks
      \n(xii) Glucose
      \n(xiii) Petrol\/Diesel\/Kerosene oil
      \n(xiv) Steam
      \n(xv) Cloud.
      \nAnswer:
      \nElement – (iv), (ix); Compounds – (viii), (xii), (xiv), (xv); Mixtures – (i), (ii), (iii), (v), (vi), (vii), (x), (xi), (xiii).<\/p>\n

      Question 30.
      \nWhy is air sometimes considered as a heterogeneous mixture?
      \nAnswer:
      \nThis is due to the presence of dust particles which change the phase, into
      \nheterogeneous.<\/p>\n

      \"KSEEB<\/p>\n

      Question 31.
      \n1 L of a gas at STP weighs 1.97 g. What is the vapour density of the gas?
      \nAnswer:
      \n22.4 L of the gas at STP will weigh 1.97 \u00d7 22.4 = 44.1 g ie., molecular mass = 44.1.
      \n\"1st<\/p>\n

      Question 32.
      \nWhy atomic masses are the average values ?
      \nAnswer:
      \nMost of the elements exist in different isotopes ie., atoms with different masses, eg. Cl has two isotopes with mass numbers 35 and 37 existing in the ratio 3:1. Hence, average value is taken.<\/p>\n

      Question 33.
      \nDetermine the equivalent weight of each of the following compounds assuming the formula weights of these compounds are x, y and z respectively, (i) Na2<\/sub>SO4 <\/sub>(ii) Na3<\/sub>P04<\/sub>.12H2<\/sub>O (iii) Ca3<\/sub>(PO4<\/sub>)2<\/sub>
      \nAnswer:
      \n\"1st
      \nQuestion 34.
      \nWhat is the mass of mercury in grams and in kilograms if the density of liquid mercury is 13.6 g cm-3<\/sup>?
      \nAnswer:
      \nMass = Volumexdensity = 1000 cm3<\/sup> \u00d7 13.6 g cm-3<\/sup> =13,600 g = 13.6 kg<\/p>\n

      Question 35.
      \nVitamin C is known to contain 1.29 \u00d7 1024<\/sup> hydrogen atoms. Calculate number of moles of hydrogen atoms.
      \nAnswer:
      \n\"1st<\/p>\n

      Question 36.
      \nDefine an organic compound. Give one example.
      \nAnswer:
      \nOrganic compound contains essentially carbon, hydrogen along certain elements like N, P, S, O, halogen, etc., eg: urea, sugar.<\/p>\n

      Question 37.
      \nDefine an inorganic compound. Give one example.
      \nAnswer:
      \nInorganic compound is made up of two or more elements containing rare percentage of carbon eg: NaCl, KC1, ZnS, etc.<\/p>\n

      Question 38.
      \nName the process used for refining of petroleum.
      \nAnswer:
      \nFractional distillation.<\/p>\n

      Question 39.
      \nCalculate the percentage of nitrogen in NH4<\/sub>. (Atomic mass of N = 14, H = 1 amu)
      \nAnswer:
      \n\"1st<\/p>\n

      Question 40.
      \nState the law of multiple proportion.
      \nAnswer:
      \nIt states \u2018whenever two elements combine to form two or more compounds, the ratio between different weights of one of the elements which combines with fixed weight of other is always simple’.<\/p>\n

      \"KSEEB<\/p>\n

      Question 41.
      \nCalculate the number of He atoms in (i) 52 u, (ii) 52 g, (iii) 52 moles of He. Atomic weight of He is 4 u.
      \nAnswer:
      \n(i) Atomic weight of Helium = 4, i.e. 4 u is mass of 1 atom,
      \n52 u is mass of \\(\\frac { 1 }{ 4 }\\) \u00d7 52 = 13 atoms.<\/p>\n

      (ii) 4 g of He contains 6.022 \u00d7 1023<\/sup> He atoms \/ 52 g of He contains ?
      \n\\(\\frac{6.022 \\times 10^{23} \\times 52}{4}\\) = 7.8286 \u00d7 1023<\/sup> atoms.<\/p>\n

      (iii) 1 mole of He contains 6.022 \u00d7 1023<\/sup> atoms, 52 moles of He contains 52 \u00d7 6.022 \u00d7 1023<\/sup> =3.131 \u00d7 1023<\/sup> atoms.<\/p>\n

      Question 42.
      \nHow many electrons are present in 16 g of CH4<\/sub> ?
      \nAnswer:
      \n1 molecule of CH4<\/sub> =6 + 4 = 10 electrons, 16 g of CH4<\/sub> contains 10 \u00d7 6.022 \u00d7 1023<\/sup> = 6.022 \u00d7 1024<\/sup><\/p>\n

      Question 43.
      \nBoron occurs in nature in the form of two isotopes and \\(\\begin{array}{l}{11} \\\\ {5}\\end{array}\\) B and \\(\\begin{array}{l}{10} \\\\ {5}\\end{array}\\) in ratio of 81% and 19% respectively. Calculate its average atomic mass.
      \nAnswer:
      \n\"1st<\/p>\n

      Question 44.
      \nIf 2 litres of N2<\/sub> is mixed with 2 litres of H2<\/sub> at a constant temperature and pressure. What will be the volume of NH3<\/sub> formed?
      \nAnswer:
      \nN2<\/sub>(g) + 3H2<\/sub>(g) \u2192 2NH3<\/sub>(g)<\/p>\n

      \"KSEEB<\/p>\n

      Question 45.
      \nHow many atoms are present in 4 ml of NH3 at STP?
      \nAnswer:
      \n22400 ml of NH3<\/sub> contains = 4 \u00d7 6.022 \u00d7 1023<\/sup> atoms [\u2235 NH3<\/sub> contains 4 atoms]. 1ml of NH3<\/sub> contains = \\(\\frac{4 \\times 6.022 \\times 10^{23}}{22400}\\) = 1.07 \u00d7 1020<\/sup> atoms.<\/p>\n

      Question 46.
      \nWhich of these weighs most? (i) 32 g of oxygen, (ii) 2 g atom of hydrogen, (iii) 0.5 mole of Fe, (iv) 3.01 \u00d7 1023<\/sup> atoms of carbon.
      \nAnswer:
      \n(i) 32 g of oxygen weighs most.
      \n(ii) 2 g atom of H2<\/sub> = 2g
      \n(iii) 0.5 mole of Fe = mole \u00d7 atomic weight = 0.5 \u00d7 56 = 28 g
      \n(iv) 6.022 \u00d7 1023<\/sup> = 1 atom of d 3.01 \u00d7 1023<\/sup> atoms of C = \\(\\frac { 1 }{ 2 }\\) \u00d7 12 = 6 g.<\/p>\n

      Question 47.
      \nAn element has a specific heat of 0.113 cal\/g \u00b0C. Calculate atomic weight of element.
      \nAnswer:
      \nAccording to Dulong Petitt’s law, Atomic weight \u00d7 Specific heat = 6.4 cal
      \nAtomic weight \u00d7 0.113 cal\/g = 6.4 cal
      \nAtomic weight = \\(\\frac { 6.4 }{ 0.113 }\\) = 56<\/p>\n

      \"KSEEB<\/p>\n

      Question 48.
      \nWhy are the atomic masses of most of the elements fractional?
      \nAnswer:
      \nThis is because atomic masses are the relative masses of atoms as compared with an atom of C-12 isotope taken as 12.<\/p>\n

      Question 49.
      \nMatch the following:
      \n\"1st
      \nAnswer:
      \nA-(ii)
      \nB-(iii)
      \nC-(iv)
      \nD-(i)<\/p>\n

      Question 50.
      \nMatch the following prefixes with their multiplies.
      \n\"1st
      \nAnswer:
      \n(i) micro 10-6<\/sup>
      \n(ii) deca – 10
      \n(ill) mega – 106<\/sup>
      \n(iv) giga – 109<\/sup>
      \n(v) femto – 10-15<\/sup><\/p>\n

      Question 51.
      \nCalculate the molar mass of the following:
      \n(i) H2<\/sub>O (ii) CO2<\/sub> (iii) CH4<\/sub>
      \nAnswer:
      \n(i) H2<\/sub>O = 2 \u00d7 H+ 1 \u00d7 O
      \n2 \u00d7 1 + 1 \u00d7 16 = 18
      \n(ii) CO2<\/sub> = 12 \u00d7 1 + 2 \u00d7 16 = 44
      \n(iii) CH4 = 1 \u00d7 C + 4 \u00d7 H
      \n1 \u00d7 12 + 4 \u00d7 1 =16<\/p>\n

      1st PUC Chemistry Some Basic Concepts of Chemistry Two Marks Questions and Answers<\/h3>\n

      Question 1.
      \nDefine basicity of an acid with example.
      \nAnswer:
      \nBasicity of an acid is the number of replaceable hydrogen atoms present on a molecule of an acid is called basicity of an acid.
      \n\"1st<\/p>\n

      \"KSEEB<\/p>\n

      Question 2.
      \nDefine acidity of a base with example.
      \nAnswer:
      \nAcidity of a base is the number of a mono basic acid required to neutralize 1 molecule of a base. [The base which contain number of replaceable – OH group].
      \n\"1st<\/p>\n

      Question 3.
      \nDefine equivalent weight of acid with example.
      \nAnswer:
      \nThe number of parts by mass of the acid which contains 1 part by mass of replaceable hydrogen.
      \n\"1st<\/p>\n

      Question 4.
      \nCalculate equivalent weight of following acid, (a) H2<\/sub>C2<\/sub>O4<\/sub>.H2<\/sub>O (b) H2<\/sub>O2<\/sub>O4
      \n<\/sub>Answer:
      \n\"1st<\/p>\n

      Question 5.
      \nDefine equivalent weight of base with an example.
      \nAnswer:
      \nNaOH + HCl \u2192 NaCl + H2<\/sub>O
      \n1 equivalent of HC1 = 36.45, here 1 equivalent of HCl reacts with NaOH.
      \n\"1st<\/p>\n

      \"KSEEB<\/p>\n

      Question 6.
      \nWrite the acidity and calculate equivalent weight of following base, a) Ca(OH)2<\/sub> b) Fe(OH)3<\/sub>
      \nAnswer:
      \n\"1st<\/p>\n

      Question 7.
      \nWrite the mathematical form for mole fraction.
      \nAnswer:
      \nMathematically;
      \nIf n1<\/sub> is the number of moles of solvent; If n2<\/sub> is the number of moles of solute
      \n\"1st<\/p>\n

      Question 8.
      \nWhat is ppm (parts per million)?
      \nAnswer:
      \nIt is the mass of solute present in 1 million(104<\/sup>) of solution.
      \n\"1st
      \n10-3<\/sup> = 1 mg of solute \/1 dm3<\/sup> of solution.
      \n\u2234 1 ppm = 1 mg of solute \/1 dm3<\/sup> of solution.
      \nn ppm = n mg solute \/ 1 dm3<\/sup> of solution.<\/p>\n

      Question 9.
      \nDefine Normality.
      \nAnswer:
      \nNumber of gram equivalent of the substance dissolved in 1 dm3<\/sup> of solution.
      \n\"1st<\/p>\n

      Question 10.
      \nCalculate molarity and normality of the following solution, a) HCl b) H2<\/sub>C2<\/sub>O4<\/sub>
      \nAnswer:
      \n\"1st<\/p>\n

      Question 11.
      \nWrite a note on internal indicator.
      \nAnswer:
      \nInternal indicator are the substance which are added in the titration flask at the beginning. Example: Methyl orange, starch, phenolphthalein, diphenylamine.
      \nInternal indicator are further divided according to the type of reaction taking place during titration.<\/p>\n

      (a) Acid base indicator: The choice of indicator for acid base titration depends on the nature of the acid and base used in that reaction.
      \nAcid Base Indicator
      \nStrong acid Strong base Methyl orange or phenolphthalein
      \n\"1st
      \n(b) Redox indicator: Example: Diphenylamine. When titrating KMnO4<\/sub> with oxalic acid.<\/p>\n

      Question 12.
      \nWhat are external indicator? Give an example.
      \nAnswer:
      \nExternal indicator is added in between the titration process by taking a drop of titrated
      \nmixture.
      \nExample: Potassium ferrocynide is used when ferrous ammonium sulphate is titrated against K2<\/sub>Cr2<\/sub>O7<\/sub>.<\/p>\n

      Question 13.
      \nWhat is self indicator ? Give an Example.
      \nAnswer:
      \nWhen in titration one of the solution acts as a self indicator. Example: KMnO4<\/sub>.<\/p>\n

      \"KSEEB<\/p>\n

      Question 14.
      \nWhat is gram molecular value? Explain.
      \nAnswer:
      \nThe volume occupied by one gram mole of a substance in the gaseous or is the vapour state at STP is called the gram molecular volume. The gram molecular volume of all the substance is found to be 22.4 dm 3<\/sup> or 22400 cm 3<\/sup>. The mass in gram of 22.4 dm 3<\/sup> of a gas or vapour at STP is equal to its gram molecular weight.<\/p>\n

      1 g molecule of chlorine or 1 gram molecule of carbon di-oxide or 1 gram of any other substance in the gaseous state occupies a volume of 22400 cm3<\/sup> (22.4 dm3<\/sup> ).<\/p>\n

      1 g molecule weight of hydrogen = 2.016 g, 2.016 g of hydrogen occupies 22.4 dm3<\/sup> of volume at STP. But 1 g equivalent weight of hydrogen is 1.008 g occupies 11.2 dm3<\/sup> of volume at STP or mass of 11200 cm 3<\/sup> of hydrogen = its equivalent weight.<\/p>\n

      Question 15.
      \nDefine the following with example, (i) Allotropy and allotropes, (ii) polymorphism, (iii) isomorphism.
      \nAnswer:
      \n(i) The existence of an element in two or more chemically similar but physically different forms is called allotropy and the different forms are called allotropes, e.g., diamond, graphite, wood charcoal, lamp black etc., are allotropes of carbon.
      \n(ii) The existence of a compound in different crystalline forms is called polymorphism and the different forms are called polymorphs, e.g., ZnS has two polymorphs called zinc blende and wurtzite.
      \n(iii) The existence of different compounds with similar chemical composition in the same crystalline form is called [M2<\/sub>SO4<\/sub>M2<\/sub> (SO4<\/sub>)3<\/sub> 24H2<\/sub>O] are isomorphs.<\/p>\n

      Question 16.
      \nDiscuss smaller the quantity to be measured more precise should be the instrument.
      \nAnswer:
      \nQuite often the uncertainty in measurement is expressed in terms of percentage by putting \u00b1 sign before it, e.g., 250 \u00b1 1% etc. If the same instrument is used for measuring different quantities, then smaller the quantity to equal to \u00b1 1 mg, then if we weigh 100 g on it, the result can be reported as 100 \u00b1 0.001%. If same balance is used to weigh 10 g, the result reported will be 10 \u00b1 0.01%, and if 1 g is weighed, the result reported will be 10 \u00b1 0.1%. Hence, smaller the quantity to be measured, more precise should be the instrument.<\/p>\n

      Question 17.
      \nCompute the mass of one molecule and the molecular mass of C6<\/sub>H6<\/sub> \u00a0(benzene). (At. mass of C = 12, H = 1 u).
      \nAnswer:
      \nMolecular weight of C6<\/sub>H6<\/sub> = 6 \u00d7 12 + 6 \u00d7 1 = 78 g,
      \nMass of 1 molecule = \\(\\frac{78}{6.002 \\times 10^{23}}\\) = 12.94 \u00d7 10-23<\/sup> g = 1.294 \u00d7 10-22<\/sup> g<\/p>\n

      Question 18.
      \nAn organometallic compound on analysis was found to contain, C = 64.4%, H = 5.5% and Fe = 29.9%. Determine its empirical formula. (At. mass of Fe = 56 u).
      \nAnswer:
      \n\"1st<\/p>\n

      Question 19.
      \n4g of copper chloride on analysis was founded to contain 1.890 g of copper (Cu) and 2.110 g of chlorine (Cl). What is the empirical formula of copper chloride? [At. mass of Cu = 63.5 u, Cl = 35.5 u].
      \nAnswer:
      \n\"1st<\/p>\n

      \"KSEEB<\/p>\n

      Question 20.
      \nCalculate the number of grams of oxygen in 0.10 mol of Na2<\/sub>CO3<\/sub>.10H2<\/sub>O.
      \nAnswer:
      \n1 mole of Na2<\/sub>CO3<\/sub>.10H2<\/sub>O contains 13 moles of oxygen atoms.
      \n0.1 mole Na2<\/sub>CO3<\/sub>.10H2<\/sub>O contains 13 \u00d7 0.1 = 1.3 moles of oxygen atoms.
      \nMass of oxygen atoms = 1.3 \u00d7 16 = 30.8 g.<\/p>\n

      Question 21.
      \nHow many grams of Cl2 are required to completely react with 0.4 g of H2 to yield HC1? Also calculate the amount of HC1 formed.
      \nAnswer:
      \nH2(g)<\/sub> + Cl2(g)<\/sub> \u2192 HCl(g)<\/sub> ,2 \u00d7 1 = 2 2 \u00d7 35.5 = 71
      \n2 g of H2<\/sub> reacts with 71 g of Cl2<\/sub> to give 73 g of HCl.
      \n0.4 g of H2<\/sub> reacts with \\(\\frac{71}{2}\\)x 0.4 = 14.2 g of Cl2<\/sub> to give \\(\\frac{73 \\times 0.4}{2}\\) = 14.6 g of HCl.
      \n14.2 g of Cl2<\/sub> and 14.6 g of HCl.<\/p>\n

      Question 22.
      \nCone. HC1 is 38% HCl by mass. What is the molarity of this solution if d = 1.19g cm-3<\/sup>? What volume of cone. HC1 is required to make 1.00 L of 0.10 M HCl?
      \nAnswer:
      \n\"1st<\/p>\n

      Question 23.
      \n\"1st
      \nAnswer:
      \n\"1st<\/p>\n

      Question 24.
      \nCalculate the volume of O2 <\/sub>at STP liberated by heating 12.25 g of KClO3<\/sub>. (At. wt. of K = 39, Cl = 35.5, O = 16 u).
      \nAnswer:
      \n\"1st<\/p>\n

      Question 25.
      \n1 M solution of NaNO3<\/sub> has density 1.25 g cm -3<\/sup>. Calculate its molality. (Mol. weight of NaNO3<\/sub> = 85 g mol-1<\/sup>).
      \nAnswer:
      \nMass of solution = Volume of solution \u00d7 density of solution
      \nMass of solution = 1000 cm3<\/sup> \u00d7 1.25 g cm-3<\/sup> = 1250 g
      \nMass of solute = 85 g; Mass of solvent = 1250 – 85 = 1165 g.
      \n\"1st<\/p>\n

      \"KSEEB<\/p>\n

      Question 26.
      \nExplain how compounds differ from elements? Give two differences.
      \nAnswer:
      \nElements consist of only one kind of atoms.
      \n\"1st<\/p>\n

      Question 27.
      \nExplain how mixture differs from pure substance? Give 2 difference.
      \nAnswer:
      \nPure substances Mixtures
      \n\"1st<\/p>\n

      Question 28.
      \nClassify each of the following as pure substance or mixture: (a) Ethyl alcohol, (b) Oxygen, (c) Blood, (d) Carbon, (e) Steel, (f) Distilled water.
      \nAnswer:
      \nPure substances: (a) Ethyl alcohol, (b) Oxygen, (d) Carbon, (f) Distilled water. Mixtures: (c) Blood and (e) Steel.<\/p>\n

      Question 29.
      \nDescribe two factors that introduce uncertainty into measured figures.
      \nAnswer:
      \n(i) Reliability of measuring instrument
      \n(ii) Skill of the person making the measurement.<\/p>\n

      Question 30.
      \nA 50 ml graduated cylinder has 1 ml graduation. What is maximum number of significant figures of the volumes that can be reported from this graduated cylinder?
      \nAnswer:
      \nThe maximum number of significant figures of the volumes that can be reported are 2.<\/p>\n

      Question 31.
      \nWhat is the difference between accuracy and precision of measurements?
      \nAnswer:
      \nAccuracy: It is related to closeness of signal measurement to its true value.
      \nPrecision: It refers to the closeness of the set of values obtained from identical measurements of a quantity.<\/p>\n

      Question 32.
      \nClassify the following substances into elements, compounds, and mixture.
      \n(1) Milk (2) Iodised table salt (3) Diamond (4) Steel (5) Dry Ice (6) Glucose (7) Petrol (8) Steam (9) Cloud (10) Aerated drinks
      \nAnswer:
      \nElements (3), Compounds 5,6, 8,9, Mixture 1, 2,4, 7,10<\/p>\n

      \"KSEEB<\/p>\n

      1st PUC Chemistry Some Basic Concepts of Chemistry Five Marks Questions and Answers<\/h3>\n

      Question 1.
      \nCalculate the moles of NaOH required to neutralize the solution produced by dissolving 1.1 g P4<\/sub>O6<\/sub> in water. Use the following reactions:
      \nP4<\/sub>O6<\/sub> + 6H2<\/sub>O \u2192 4H3<\/sub>PO3<\/sub>;
      \n2NaOH + H3<\/sub>PO3<\/sub> \u2192 Na2<\/sub>HPO3<\/sub> + 2H2<\/sub>O
      \n(At. mass\/g mol-1<\/sup> P = 31, O = 16).
      \nAnswer:
      \nMolecular weight of P4<\/sub>O6<\/sub> =4 \u00d7 31 + 6 \u00d7 16 =124 + 96 = 220 gmol-1<\/sup>
      \n\"1st
      \n\\(\\frac { 1 }{ 200 }\\) moles of PiOfi produce 4 \u00d7 \\(\\frac { 1 }{ 200 }\\)moles of H3<\/sub>PO3<\/sub>
      \n1 mole of H3<\/sub>PO3<\/sub> requires 2 moles of NaOH
      \n\\(\\frac { 1 }{ 50 }\\)moles of H3<\/sub>PO3<\/sub> requires 2 \u00d7 \\(\\frac { 1 }{ 50 }\\) = \\(\\frac { 1 }{ 25 }\\) = 0.04 moles of NaOH.<\/p>\n

      Question 2.
      \n(a) A sample of NaOH weighing 0.38 g is dissolved in water and the solution is made to 50.0 cm3<\/sup> in a volumetric flask. What is the molarity of the solution?
      \n(b) State and explain law of multiple proportion.
      \nAnswer:
      \n\"1st<\/p>\n

      (b) Law of multiple proportion states whenever two elements react to form two or more compounds, the ratio between different weights of one of the elements which combine with fixed weight of another is always simple, e.g., nitrogen reacts with oxygen to form NO and NO2<\/sub>. In NO, 14 g of N reacts with 16 g of O. In NO2<\/sub>,14 g of N reacts with 32 g of O. The ratio between weights of oxygen which combine with fixed weight of nitrogen is 16 : 32. i.e., 1:2.<\/p>\n

      Question 3.
      \nCalculate the amount of water (g) produced by the combustion of 16 g of methane.
      \nAnswer:
      \nCH4(g) <\/sub>+ 2O2(g)<\/sub> >CO2(g) <\/sub>+ 2H2<\/sub>O(1)<\/sub>
      \n1 mole = 16 g of CH4<\/sub> gives 2 moles of H2<\/sub>O i.e., 2 \u00d7 18 = 36 g of H2<\/sub>O.<\/p>\n

      Question 4.
      \nHow many moles of methane are required to produce 22 g of CO,2(g) after combustion?
      \nAnswer:
      \n44 g of CO2<\/sub> will be produced by 1 mole of CH4<\/sub>.
      \n22 g of CO2<\/sub> will be formed by \\(\\frac{1}{44}\\) \u00d7 22 = 0.5 mole of CH4<\/sub>.<\/p>\n

      \"KSEEB<\/p>\n

      Question 5.
      \n(a) How many significant figures are there in 1.00 \u00d7 106<\/sup>?
      \n(b) One mole of sugar contains ……. oxygen atoms.
      \n(c) Give an example of molecule in which the empirical formula is CH2<\/sub>O and the ratio of molecular formula weight and empirical formula weight is 6.
      \nAnswer:
      \n(a) 3
      \n(b) One mole of sugar C12<\/sub>H22<\/sub>O11<\/sub> contains 11 \u00d7 6.023 \u00d7 1023<\/sup> oxygen atoms.
      \n= 66.253 \u00d7 10 23<\/sup> = 6.6253 \u00d7 10 23<\/sup> atoms of oxygen
      \n(c) C6<\/sub>H12<\/sub>O6<\/sub>.<\/p>\n

      Question 6.
      \nAn organic monobasic acid was found to contain 39.5% carbon, 6.4% hydrogen and the rest oxygen. If the equivalent mass of the acid is 60, find out its molecular formula.
      \nAnswer:
      \nPercentage of oxygen = 100 – (39.5 + 6.4) = 100 – 45.9 – 54.1
      \n\"1st
      \n\u2234Empirical formula = CH2<\/sub>O
      \nMolecular mass = Eq. mass \u00d7 basicity = 60 \u00d7 1 = 60
      \nEmpirical formula weight = CH20 = 12 + 2 + 16 = 30
      \nNows 30n = 60 => n = 2
      \n\u2234Molecular formula = (Empirical formula) \u00d7 n = (CH2<\/sub>O)2<\/sub> = C2<\/sub>H4<\/sub>O2<\/sub><\/p>\n

      Question 7.
      \nCalculate the mass of 95% pure Mn02 to produce 35.5 g of Cl2<\/sub> as per the following reaction. Mn02 + 4HC1 \u00bbMnCl2 + Cl2 + HaO.
      \n(At. mass of Mn = 55)
      \nAnswer:
      \n\"1st
      \n71 g of Cl2<\/sub> is produced by 87 g of MnO2<\/sub>
      \n35.5 g of Cl2<\/sub> is produced by \\(\\frac { 87 }{ 71 }\\) \u00d7 35.5 = 43.49 g of MnO2<\/sub>
      \nSince MnO2<\/sub> is 95% pure i.e., 95 g of MnO2<\/sub> is present in 100 g of MnO2<\/sub> sample.
      \n43.49 of pure MnO2<\/sub> will be present in \\(\\frac{43.49 \\times 100}{95}\\)= 45.8f of MnO2<\/sub> sample<\/p>\n

      Question 8.
      \n500 ml of Na2<\/sub>CO3<\/sub> solution contains 2.65 g of Na2<\/sub>CO3<\/sub> (mol. mass of Na2<\/sub>CO3<\/sub> = 106). If 10 ml of this solution is diluted to 1 L, what is the concentration of the resultant solution?
      \nAnswer:
      \nAnswer:
      \n\"1st
      \nFor dilution M1<\/sub>V1<\/sub> = M2<\/sub>V2<\/sub> \u21d2 0.05 \u00d7 10 = 1000 \u00d7 M2<\/sub>
      \nM2<\/sub> = 0.0005 M (Molarity of the resultant solution).<\/p>\n

      \"KSEEB<\/p>\n

      Question 9.
      \n2 L of a solution is prepared by dissolving 0.02 mole of NaBr and 0.02 mole of Na2<\/sub>SO4<\/sub>.1 L of mixture X + excess of AgNO3<\/sub> \u2192 Y (pale yellow ppt.).
      \n1 L of mixture X + excess of BaCl2 \u2192 Z (white ppt.).
      \nWhat is the ratio of number of moles of Y and Z?
      \nAnswer:
      \nAgNO3<\/sub> reacts with Br ions to form AgBr as pale yellow ppt. as per the following reaction
      \n\"1st
      \n2 L of solution will contain 0.02 mole of NaBr or 0.02 mole of Br–<\/sup>.
      \n=> 1 L of solution will contain 0.01 mole of Br–<\/sup>
      \n\u2234 as per equation (i) number of moles of AgBr i.e., Y formed is 0.01 mole
      \nBaCl2<\/sub> reacts with SO4<\/sub>2-<\/sup> ions present in the solution as per the following reaction
      \n\"1st
      \n2L of solution contain 0.02 mole of Na2<\/sub>SO4<\/sub> or 0.02 mole of SO4<\/sub>2-<\/sup>.
      \n\u21d2 1 L of solution will contain 0.01 mole of SO4<\/sub>2-<\/sup>.
      \n\u2234 as per equation (ii) number of moles of BaSO4<\/sub> i.e., Z formed is 0.01 mole
      \nRatio of Y and Z = 0.01: 0.01 = 1:1<\/p>\n

      Question 10.
      \nHow many grams of copper will get replaced in 2 L of CuSO4<\/sub> solution having molarity 1.50 M, if it is made to react with 54 g of aluminium? (At. mass of Cu = 63.5 and A1 = 27.0)
      \nAnswer:
      \nThe reaction with CuSO4<\/sub> solution and aluminium may be represented as
      \n3CUSO4<\/sub> + 2Al \u2192 Al2<\/sub>(SO4<\/sub>)2<\/sub> +3CU
      \n(3 \u00d7 159.5)g = 478.5g (2 \u00d7 27)g = 54g
      \nCalculation of wt. of CuSO4<\/sub> in 2 L of CuSO4<\/sub> solution having 1.50 molarity
      \n\"1st
      \n478.5 g of CuSO4<\/sub> reacts with 54 g of aluminium
      \nAlso, 159.5 g of CuSO4<\/sub> has 63.5 g of Cu
      \n1 g of CuSO4<\/sub> has \\(\\frac{63.5}{159.5}\\) g of Cu
      \n478.5 g of CuSO4<\/sub> has \\(\\frac{63.5}{159.5}\\) \u00d7 478.5g of Cu = 190.5g of Cu<\/p>\n

      Question 11.
      \n0.7 g of Na2<\/sub>CO3<\/sub>.xH2<\/sub>O dissolved to make 100 ml of solution. If 20 ml of this solution is required to completely neutralize 19.8 ml of 0.1 M HCl, then what is the value of x?
      \nAnswer:
      \nCalculation of molarity of sodium carbonate solution
      \nn1<\/sub>M1<\/sub>V1<\/sub> = n2<\/sub>M2<\/sub>V2<\/sub> (n factor for sodium carbonate is 2)
      \n2 \u00d7 20 \u00d7 M1<\/sub> =19.8 \u00d7 0.1 \u21d2 M1<\/sub> =0.0495
      \n\"1st
      \nThe value of x i.e., number of water molecules of crystallization is 2.<\/p>\n

      Question 12.
      \nCalculate the weight of CaO that can be obtained by heating 200 kg of limestone which is 95% pure. (At. mass of Ca = 40, C = 12 and O = 16).
      \nAnswer:
      \n\"1st
      \n100 kg of CaCO3<\/sub> gives 56 kg of CaO.
      \nSince lime stone is 95% pure; amount of CaCO3<\/sub> present in 200 kg of impure sample = 190 kg CaCO3<\/sub> 100 kg of CaCO3<\/sub> gives 56 kg of CaO
      \n1 kg of CaCO3<\/sub> gives \\(\\frac { 56 }{ 100 }\\)kg of CaO
      \n190 kg of CaCO3<\/sub> give \\(\\frac { 56 }{ 100 }\\) \u00d7 190 = 106.4 kg of CaO.<\/p>\n

      \"KSEEB<\/p>\n

      Question 13.
      \n20 g of sample containing Ba(OH)2<\/sub> is dissolved in 10 ml of 0.5 M HC1 solution. The excess of HCl was then titrated against 0.2 M NaOH. The volume of NaOH used in the titration was 10 ml. Calculate the percentage of Ba(OH)2<\/sub> in the sample. (Mol. wt. of Ba(OH)2<\/sub> = 171)
      \nAnswer:
      \nCalculation of volume of HCl used in titration between NaOH and HCl.
      \nVNaOH<\/sub> = 10 ml, MNaOH<\/sub> = 0.2M ,MHCl<\/sub> = 0.5M, VHCl<\/sub> = ?
      \nMNaOH<\/sub> \u00d7 VNaOH<\/sub>=MHCl<\/sub> \u00d7 VHCl<\/sub> VHCl<\/sub> = \\(\\frac{10 \\times 0.2}{0.5}\\) = 4ml
      \nThis is the volume of HCl left unused when excess of HC1 is added to Ba(OH)2<\/sub> solution.
      \nTotal volume of HCl added = 10 ml.
      \nVolume of HCl used to react with Ba(OH)2<\/sub> = 10 – 4 = 6 ml
      \n\u2234HCl react with Ba(OH)2<\/sub> as per the equation
      \n2HCl + Ba(OH)2<\/sub> \u2192 BaCl2<\/sub> + 2H2<\/sub>O
      \n6 ml of 0.5 M HCl has reacted with Ba(OH)2<\/sub> => Number of moles of HCl reacted,
      \n\"1st
      \nObserving the molar ratio of HC1 and Ba(OH)2<\/sub>,
      \nMoles of Ba(OH)2<\/sub> reacted = \\(\\frac { 1 }{ 2 }\\) \u00d7 moles of HCl reacted = \\(\\frac { 1 }{ 2 }\\) \u00d7 O.003 = 0.0015 moles
      \nWeight of Ba(OH)2<\/sub> reacted = no. of moles \u00d7 mol. wt. = 0.0015\u00d7 171 = 0.2565 g
      \n\"1st<\/p>\n

      1st PUC Chemistry Some Basic Concepts of Chemistry\u00a0Numerical Problems and Answers<\/h3>\n

      Question 1.
      \nHow many significant figures are there in each of the following numbers?
      \n(i) 6.005, (ii) 6.002 \u00d7 1023<\/sup>, (iii) 8000, (iv) 0.0025, (v) \u03c0, (vi) the sum 18.5 + 0.4235, (vii) the product 14 \u00d7 6.345.
      \nAnswer:
      \n(i) Four because the zeros between the non-zero digits are significant figures.
      \n(ii) Four because only the first term gives the significant figures and exponential term is not considered.
      \n(iii) Four. However, if expressed in scientific notation as 8 \u00d7 103<\/sup> , it will have only one significant figure, as 8.0 \u00d7 103<\/sup>, 8.00 \u00d7 103<\/sup> or 8.000 x 103<\/sup>, it will have 2, 3 or 4 significant figures.
      \n(iv) Two because the zeros on the left of the first non-zero digit are not significant.
      \n(v) As \u03c0 = \\(\\frac { 22 }{ 7 }\\) = 3.1428571…., hence it has infinite number of significant figures.
      \n(vi) Three, because the reported sum will be only upto one decimal place i.e., 18.9.
      \n(vii) Two, because the number with least number of significant figures involved in the calculation (i.e., 14) has two significant figures.<\/p>\n

      Question 2.
      \nExpress the following to four significant figures: (i) 6.45372, (ii) 48.38250, (iii) 70000, (iv) 2.65986 \u00d7 103<\/sup>, (v) 0.004687.
      \nAnswer:
      \n(i) 6.454, (ii) 48.38, (iii) 7.000 \u00d7 104, (iv) 2.660 \u00d7 103<\/sup>, (v) 0.004687.<\/p>\n

      Question 3.
      \nA sample of nickel weighs 6.5425 g and has a density of 8.8 g cm3<\/sup> the volume? Report the answer to correct decimal place.
      \nanswer:
      \n\"1st
      \nThe result should have two significant figures because the least precise term (8.8) has two significant figures.<\/p>\n

      \"KSEEB<\/p>\n

      Question 4.
      \nExpress the result of the following calculation to the appropriate number of \\(\\frac{3.24 \\times 0.08666}{5.006}\\)
      \nAnswer:
      \n\\(\\frac{3.24 \\times 0.08666}{5.006}\\)= 0.0560883 (Actual result)
      \nAs 3.24 has least number of significant figures, viz., 3 , the result should contain 3 significant figures only. Hence, the result will be reported as 0.0561 (after rounding off).<\/p>\n

      Question 5.
      \nThe mass of precious stones is at pressed in terms of carat. Given that 1 carat = 3.168 grains and 1 gram – 15.4 grains, calculate the total mass of the ring in grams and kilograms which contains 0.500 carat diamond and 7.00 gram gold.
      \nAnswer:
      \nThe unit conversion factors to be used will be;
      \n\"1st<\/p>\n

      Question 6.
      \nTwo oxides of a metal contain 27.6% and 30.0% of oxygen respectively. If the formula of the first oxide is M3<\/sub>O4<\/sub>, find that of the second.
      \nAnswer:
      \nIn the first oxide, oxygen = 27.6, metal = 100 – 27.6 = 72.4 parts by mass.
      \nAs the formula of the oxide is M3<\/sub>O4<\/sub>, this means 72.4 parts by mass of metal = 3 atoms of metal and 4 atoms of oxygen = 27.6 parts by mass.
      \nIn the second oxide, oxygen = 30.0 parts by mass and metal = 100 – 30 = 70 parts by mass.
      \nBut 72.4 parts by mass of metal = 3 atoms of metal.
      \n\u2234 70 parts by mass of metal =\\(\\frac{3}{72.4}\\) \u00d7 70 atoms of metal = 2.90 atoms of metal.
      \nAlso, 27.6 parts by mass of oxygen = 4 atoms of oxygen.
      \n\u2234 4.35 atoms of oxygen.
      \nHence, ratio of M: O in the second oxide = 2.90 : 4.35 = 1:1.5 = 2:3.
      \n\u2234 Formula of the metal oxide is M2<\/sub>O3<\/sub>.<\/p>\n

      Question 7.
      \nWhat is the equivalent weight of KH(IO3<\/sub>)2<\/sub> as an oxidant in presence of 4.0 (N) HCl when IC becomes the reduced form? (K = 39.0,1 = 127.0).
      \nAnswer:
      \nIn KH(IO3<\/sub>)2<\/sub>, IO3<\/sub> is present as IO3<\/sub>–<\/sup>. Oxidation state of I will be
      \n\"1st
      \nQuestion 8.
      \nCalculate the percentage of the naturally occurring isotopes 35<\/sup>C1 and 37<\/sup>Cl that accounts for the atomic mass of chlorine taken as 35.45.
      \nAnswer:
      \nSuppose 35<\/sup>Cl present = x%. Then 37<\/sup>Cl present = (100 -x)%
      \n\u2234 Average atomic mass = \\frac{x \\times 35+(100-x) \\times 37}{100} = 35.45 (Given)
      \nor 35x + 3700-37x = 3545 or 2x = 155 or x = 77.5%
      \nThis 35<\/sup>Cl = 77.5% and 37<\/sup>Cl = (100 – 77.5) = 22.5%.<\/p>\n

      Question 9.
      \nConvert 22.4 L into cubic metres.
      \nAnswer:
      \n\"1st<\/p>\n

      Question 10.
      \nCalculate the molar mass of water if it contains 50% heavy water (D2<\/sub>O).
      \nAnswer:
      \nAs water contains 50% D2<\/sub>O, this means that it contains \\(\\frac{1}{2} mole of H20 and \\frac{1}{2}\\) mole
      \nof D2<\/sub>O. Mass of \\(\\frac{1}{2}\\) mole of D2<\/sub>O = \\(\\frac{1}{2}\\) \u00d7 (2 \u00d7 2 + 16) = 10 g.
      \nHence, molar mass of the given sample of water = 9 + 10 = 19 g mol-1<\/sup>.\u00d7<\/p>\n

      Question 11.
      \n60 cc of oxygen was added to 24 cc of carbon monoxide and the mixture ignited. Calculate the volume of oxygen used up and the volume of carbon dioxide formed.
      \nAnswer:
      \n\"1st
      \n\u2234 volume of oxygen used up is 12 cc. and the volume of carbon dioxide formed is 24 cc.<\/p>\n

      \"KSEEB<\/p>\n

      Question 12.
      \n200 cm3<\/sup> of carbon monoxide is mixed with 200 cm3<\/sup> of oxygen at room temperature and ignited. Calculate the vol of CO2<\/sub> formed on cooling to room temperature. What other gas if any may also be present.
      \nAnswer:
      \n\"1st
      \nVolume of carbon dioxide formed is 200 cm3<\/sup> and the other gas present is 100 cm3<\/sup> of oxygen.<\/p>\n

      Question 13.
      \nCalculate the volume of oxygen required to burn completely a mixture of 22.4 dm3<\/sup> of CH4<\/sub> and 11.2 dm3<\/sup> of H2<\/sub> (all volumes measured at STP) [1 dm3<\/sup> = 1 litre].
      \nAnswer:
      \n\"1st
      \nTotal oxygen required = 44.8 dm3<\/sup> +5.6 dm3<\/sup> =50.4 dm3<\/sup> Pits].
      \n\u2234 the number of moles in 7 g of nitrogen is 0.25 moles.<\/p>\n

      Question 14.
      \nCalculate the mass of 50 cc of CO at STP (C = 12, O = 16).
      \nAnswer:
      \n[1 mole = 1 g mol. wt. and occupies 22.4 lit. at STP]
      \ng mol. wt. of carbon monoxide = 12 + 16 = 28 g
      \n28 g of CO occupies 22400 cc at STP
      \nHow many grams of CO will occupy 50 cc at STP = \\(\\frac{28 \\times 50}{22400}\\) = 0.0625 g<\/p>\n

      Question 15.
      \nCalculate the volume at STP occupied by 6.023 \u00d7 1022<\/sup> molecules of a gas X.
      \nAnswer:
      \n[1 mole of any substance contains 6.023 \u00d7 1023<\/sup> number of molecules [Avogadro\u2019s number]].
      \n(a) 6.022 \u00d7 1023<\/sup> molecules occupied 22.42 cm3<\/sup> (Its) = \\(\\frac{22.4 \\times 6.023 \\times 10^{22}}{6.023 \\times 10^{23}}\\) = 2.24 lits<\/p>\n

      Question 16.
      \nCalculate the number of molecules in 1 kg of sodium chloride. [Na = 23, Cl = 35.5].
      \nAnswer:
      \n(a) 1 mol wt. of NaCl = 58.5 g contains 6.022 \u00d7 1023<\/sup>
      \n(b) Contains molecules = \\(\\frac{1000}{58.5}\\) \u00d7 6.023 \u00d7 1023<\/sup> = 102.96 \u00d7 1023<\/sup> molecules.<\/p>\n

      Question 17.
      \nCalculate the atomicity of a gas X if 1 g of X occupies 11,200 cc at STP [At. Wt. of X= 1].
      \nAnswer:
      \n\"1st<\/p>\n

      Question 18.
      \n0. 48 g of a gas forms 100 cm3<\/sup> of vapours at STP. Calculate the gram molecular wt. of the gas.
      \nAnswer:
      \n100 cm3<\/sup> of the gas weighs 0.48 grams at STP
      \n22,400 cm3<\/sup> weight Mol. wt. at STP.
      \n\u2234 \\(\\frac{22400}{100}\\) \u00d7 0.48 = 107.52<\/p>\n

      \"KSEEB<\/p>\n

      Question 19.
      \nCalculate the weight of a substance X which in gaseous form occupies 10 litres at 27 \u00b0C and 700 mm pressure. The molecular Weight of X is 60.
      \nAnswer:
      \nConvert the volume to STP using the gas equation.
      \nInitial conditions: P1<\/sub> = 700 mm of Hg, V1<\/sub> = 10 litres, T1<\/sub> = 27 + 273 = 300 K
      \nFinal conditions (at STP): P0<\/sub> = 760 mm of Hg, V0<\/sub> = X litres, T0<\/sub> = 273 K
      \n\"1st<\/p>\n

      Question 20.
      \nCalculate the normality of oxalic acid solution containing 0.895 g crystals in 250 cm3<\/sup> of its solution.
      \nAnswer:
      \nMolecular formula of oxalic acid crystals = H2<\/sub>C2<\/sub>O4<\/sub>.2H2<\/sub>O = 126
      \n\"1st<\/p>\n

      Question 21.
      \n25 cm3<\/sup> of ferrous ammonium sulphate solution require 20 cm3<\/sup> of 0.1 N potassium dichromate solution. Calculate the amount of ferrous ammonium sulphate crystals dissolved in 250 cm3<\/sup> of the solution. (Given equivalent)
      \nAnswer:
      \n\"1st<\/p>\n

      Question 22.
      \nWhat should be the normality of a solution prepared by diluting 250 ml of 0.4 NH2<\/sub>SO4<\/sub> with 1000 ml of water?
      \nAnswer:
      \nTotal volume of the diluted solution = 250 + 1000 = 1250 ml
      \n\"1st<\/p>\n

      Question 23.
      \n20 cm3<\/sup> of a solution of oxalic acid requires 25 cm3<\/sup> of 0.2 N potassium . permanganate to react completely (a) Calculate the normality of oxalic acid solution, (b) What volume of this oxalic acid solution when made up to 250cm3<\/sup> gives 0.2 N solution?
      \nAnswer:
      \n\"1st
      \nNormality of oxalic acid solution =200. Volume of oxalic acid solution required =200 cm3<\/sup><\/p>\n

      Question 24.
      \n200 cm3<\/sup> of a solution of a dibasic acid contains 1.512 g of the acid and the normality of the solution is 0.12. Calculate (i) the equivalent mass and (ii) the molecular mass of the acid.
      \nAnswer:
      \n(i) Mass of the acid in one dm3<\/sup> of the solution = 1.512 \u00d7 5 = 7.56 g.
      \n\"1st
      \n(ii) Molecular mass of the acid = eq. mass \u00d7 basicity = 63 \u00d7 2 = 126.<\/p>\n

      Question 25.
      \n25 cm3<\/sup> a of a solution of sodium hydroxide required 29 cm3<\/sup> of \\(\\frac { n }{ 10 }\\) solution of oxalic acid for neutralization. Find the normality of sodium hydroxide solution and its amount dissolved in 500 cm3<\/sup> of the solution.
      \nAnswer:
      \n\"1st
      \nSubstituting we have 28 \u00d7 0.1 = 25 \u00d7 N2<\/sub>
      \n\u2234N2<\/sub> = \\(\\frac{29 \\times 0.1}{25}\\) = 0.116
      \n\u2234 Normality of sodium hydroxide solution = 0.116
      \nWe know that W = N \u00d7 E for 100 ml
      \ni.e., Mass per dm3<\/sup> = Normality \u00d7 eq. nlass = 0.116 \u00d7 40 g
      \n\u2234 Mass of sodium hydroxide in 500 cm 3 of the solution = \\(\\frac{0.116 \\times 40}{2}\\) = 2.32 g.<\/p>\n

      \"KSEEB<\/p>\n

      Question 26.
      \nCalculate the volume of concentrated nitric acid of normality 14 required to prepare 1 dm3<\/sup> of \\frac{N}{10} nitric acid.
      \nAnswer:
      \n\"1st<\/p>\n

      Question 27.
      \nCalculate the mass of hydrochloric acid in 200 cm3<\/sup> of 0.2 N solution of it. What volume of this acid solution will react exactly with 25 cm3 of 0.14 N solution of sodium hydroxide?
      \nAnswer:
      \nNormality of hydrochloric acid = 0.2; Eq. mass of hydrochloric acid = 36.5
      \nMass of hydrochloric acid in one dm3<\/sup> of the solution is given by W = N \u00d7 E = 0.2 \u00d7 36.5 g
      \n\"1st<\/p>\n

      Question 28.
      \n0. 99 g of an acid was dissolved in water and the solution made up to 200 cm3<\/sup>, 20 cm3<\/sup> of this solution required 15 cm3<\/sup> of 0.105 N sodium hydroxide solution for complete neutralization. Find the equivalent mass of the acid.
      \nAnswer:
      \n\"1st
      \nSubstituting we have 20 \u00d7 N1<\/sub> =15 \u00d7 0.105
      \n\u2234 N1<\/sub> = \\(\\frac{15 \\times 0.105}{20}\\) = 0.07875
      \n\u2234 Normality of acid solution = 0.07875
      \nMass of the acid in one dm 3 of the solution – 0.99 \u00d7 5 g = 4.95 g
      \nEq. mass of the acid is given by E = \\(\\frac{W}{N}=\\frac{4.95}{0.07875}\\) = 62.85<\/p>\n

      Question 29.
      \n2.15 g of impure oxalic acid crystals were dissolved in water and the solution made up to 250 cm3<\/sup> of this solution required 20 cm3<\/sup> of 0.16 N sodium hydroxide solution for complete neutalization. What is the percentage purity of oxalic acid crystals?
      \nAnswer:
      \n\"1st
      \nSubstituting we have 25 \u00d7 N1<\/sub> =20 \u00d7 0.16
      \n\u2234 N1<\/sub> = \\(\\frac{20 \\times 0.16}{25}\\) = 0.128
      \n\u2234 Normality of oxalic acid solution = 0.128
      \nEq. mass of oxalic acid crystals H2<\/sub>C2<\/sub>O4<\/sub>.2H2<\/sub>O = 63.
      \nMass of pure oxalic acid in one dm 3 is given by W = N \u00d7 E = 0.128 \u00d7 63
      \nMass of pure oxalic in 250 cm3 = \\(\\frac{0.128 \\times 63}{4}\\) = 2.016 g
      \nMass of impure oxalic acid crystals dissolved in 250 cm3<\/sup> = 2.15 g .
      \n% percentage purity of the given oxalic acid crystals = \\(\\frac{2.016 \\times 100}{2.15}\\) = 93.76<\/p>\n

      \"KSEEB<\/p>\n

      Question 30.
      \nA solution of sodium hydroxide contains 4.8 g of the substance per dm3, (a) What volume of the solution is required to prepare 500 cm3 of decinormal solution. (b) What volume of water has to be added to one dm3 of the solution in order to make it exactly decinormal ?
      \nAnswer:
      \n\"1st
      \nQuestion 31.
      \nFind the Molarity of hydrochloric acid containing 31.5% of hydrochloric acid. Its specific gravity is 1.16.
      \nAnswer:
      \nMass of 1000 cm3<\/sup> of the acid = 1000 \u00d7 1.16 = 1160 g
      \nMass of hydrochloric acid in 100 g of the solution = 3.15 g
      \n\"1st<\/p>\n

      Question 32.
      \n100 cm3 of concentrated hydrochloric acid which is 11 M are diluted to 275 cm3<\/sup>. If 2.5 g of limestone are required to neutralize 10 cm3<\/sup> of this diluted – acid, what is the percentage of calcium carbonate in the sample of limestone?
      \nAnswer:
      \n\"1st
      \nSubstituting we get, 100 \u00d7 11 = 275 \u00d7 M2<\/sub>
      \n\"1st
      \n\u2234 Normality of the diluted acid = 4
      \n1000 cm3<\/sup> of the acid solution contain 4 gram
      \n\u2234 10 cm3<\/sup> of the diluted acid contain = \\(\\frac{4 \\times 10}{1000}\\) i.e., 0.4 gram
      \n2.5 g of limestone contain 0.04 gram of calcium carbonate i.e., they contain 0.04 \u00d7 100 g = 4 g of calcium carbonate.
      \n(Molar mass of calcium carbonate =100)
      \n\u2234 percentage of calcium carbonate in the sample of limestone = \\(\\frac{4 \\times 100}{2.5}\\) = 16.0 .<\/p>\n

      Question 33.
      \n25.0 cm3<\/sup> of an acid required exactly 20.5 cm3<\/sup> of deci molar base for complete neutralization. What is the normality of the acid?
      \nAnswer:
      \n\"1st<\/p>\n

      Question 34.
      \nExactly 20.0 cm3<\/sup> of nitric acid neutralized 28.4 cm3<\/sup> of 0.25 M NaOH. What is the molarity of the acid?
      \nAnswer:
      \n\"1st
      \nQuestion 35.
      \n18.5 cm3<\/sup> of oxalic acid was completely neutralized by 20.0 cm3<\/sup>, 0.125 N base. Calculate the (a) normality (b) molarity and (c) mass of oxalic acid crystals in 1 dm3<\/sup> of solution.
      \nAnswer:
      \n\"1st
      \nMass of oxalic acid crystals in 1 dm3<\/sup> of solution = normality \u00d7 equivalent mass
      \n= 0.1351 N \u00d7 63.0 (equivalent mass of oxalic acid crystals = 63.0) = 8.513 g.<\/p>\n

      \"KSEEB<\/p>\n","protected":false},"excerpt":{"rendered":"

      You can Download Chapter 1 Some Basic Concepts of Chemistry Questions and Answers, Notes, 1st PUC Chemistry Question Bank with Answers Karnataka State Board Solutions help you to revise complete Syllabus and score more marks in your examinations. Karnataka 1st PUC Chemistry Question Bank Chapter 1 Some Basic Concepts of Chemistry 1st PUC Chemistry\u00a0Some Basic …<\/p>\n","protected":false},"author":5,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[81],"tags":[],"jetpack_sharing_enabled":true,"jetpack_featured_media_url":"","_links":{"self":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts\/8600"}],"collection":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/users\/5"}],"replies":[{"embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/comments?post=8600"}],"version-history":[{"count":0,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts\/8600\/revisions"}],"wp:attachment":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/media?parent=8600"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/categories?post=8600"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/tags?post=8600"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}