{"id":10159,"date":"2019-12-05T17:00:59","date_gmt":"2019-12-05T11:30:59","guid":{"rendered":"https:\/\/kseebsolutions.guru\/?p=10159"},"modified":"2021-07-02T15:43:55","modified_gmt":"2021-07-02T10:13:55","slug":"2nd-puc-chemistry-question-bank-chapter-2","status":"publish","type":"post","link":"https:\/\/kseebsolutions.guru\/2nd-puc-chemistry-question-bank-chapter-2\/","title":{"rendered":"2nd PUC Chemistry Question Bank Chapter 2 Solutions"},"content":{"rendered":"

You can Download Chapter 2 Solutions Questions and Answers, Notes, 2nd PUC Chemistry\u00a0Question Bank with Answers<\/a> Karnataka State Board Solutions help you to revise complete Syllabus and score more marks in your examinations.<\/p>\n

Karnataka 2nd PUC Chemistry Question Bank Chapter 2 Solutions<\/h2>\n
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2nd PUC Chemistry Solutions NCERT Textbook Questions and Answers<\/h3>\n

Question 1
\nDefine the term solution. How many types of solutions are formed? Write briefly about each type with an example.
\nAnswer:
\nA true solution is a homogenous mixture of two or more substances. The constituent . particle which is in larger amount\u2019’ is called a solvent and that in smaller quantity is called a solute.<\/p>\n

Since the solvent ans solute may be either . gaseous, liquid and solid, the number of possible types of binary solutions than can be prepared are given below.
\n\"2nd<\/p>\n

Question 2.
\nSuppose a solid solution is formed are very small. What kind of solid solution is this likely to be?
\nAnswer:
\nSolid in solid type. E.g: Copper in gold. This type of solutions are called alloys.<\/p>\n

Question 3.
\nDefine the following terms:
\n(i) Mole fraction
\n(ii) Molality
\n(iii) Molarity
\n(iv) Mass percentage.
\nAnswer:
\n(i) Mole fraction (X): The mole fraction of any component in a solution is the ratio of the number of moles of that component to the
\nsum of the number of moles of all the components present in the solution.
\nFor a binary solution containing A and B, Mole fraction of A,
\n\"2nd
\nwhere nA<\/sub> and nB<\/sub> are the numbers of moles of components A and B respectively.<\/p>\n

(ii) Molality (m): Molality is the numberof moles of the solute dissolved in 1000 gms (1 kg) of the solvent. It B denoted by \u2018m\u2019 mathematically.
\n\"2nd<\/p>\n

(iii) Molarity of a solution is defined as the number of moles of the solute dissolved per litre (or dm3<\/sup>) of solution. It is denoted by \u2018M\u2019 mathematically.
\n\"2nd<\/p>\n

(iv) Mass fraction multiplied by 100 gives mass percentage. E.g.: mass percentage of A
\n\"2nd<\/p>\n

\"KSEEB<\/p>\n

Question 4.
\nConcentrated nitric acid used in laboratory work is 68% nitric acid by mass in aqueous solution. What should be the molarity of such a sample of the acid if the density of the solution is 1.504 g mLr-1<\/sup>?
\nAnswer:
\n68% of nitric acid by mass means that
\nMass of nitric acid = 68g
\nMass of solution = 100g
\nMolar mass of HN03 = 63g mol-1<\/sup>?
\n\u2234 68 g HNO3<\/sub> = \\(\\frac{68}{63}\\) mole = 1.079 mole
\nDensity of solution = 1.504g mL-1<\/sup>
\n\"2nd<\/p>\n

Question 5.
\nA solution of glucose in water is labelled as 10% w\/w, what would be the molality and mole fraction of each component in the solution? If the density of solution is 1.2 g mL-1<\/sup>, then what shall be the molarity of the solution?
\nAnswer:
\n10 g glucose is present in 100 g solution, i.e., 90 g of water = 0.090 kg of H2<\/sub>O
\n\"2nd<\/p>\n

Question 6.
\nHow many mL of 0.1 M HCl are required to react completely with 1 g mixture Of Na2<\/sub>CO3<\/sub> and NaHC03 containing equimolar amounts of both?
\nAnswer:
\nStep 1: To calculate the number of moles of the components in mixture
\nsuppose Na2<\/sub>CO3<\/sub> present in the mixture = X g ,
\nNaHCO3<\/sub> present in the mixture = (1 – x) g
\nMolar mass of Na2<\/sub>CO3<\/sub> = 2 x 23+12+3 x 16=106g mol-1<\/sup>
\nMolar mass of NaHCO3<\/sub>.
\n= 23 + 1 + 12 + 3 \u00d7 16 = 84g mol-1<\/sup>
\n\u2234\u00a0 Moles of Na2<\/sub>CO3<\/sub> in xg = \\(\\frac{ x }{ 106}\\)
\nMoles of NaHCO3<\/sub> in (1-x) g = \\(\\frac{1-x}{84}\\)
\n\"2nd<\/p>\n

Step 2: To calculate the moles of HC1 required.
\nNa2<\/sub>CO3<\/sub> + 2 HCl \u2192 2 NaCl + H2<\/sub>O + CO2<\/sub>
\nNaHCO3 <\/sub>+ HCl \u2192 NaCl + H2<\/sub>O + CO2<\/sub>
\n1 mole of Na2<\/sub>CO3<\/sub> required HCl = 2 moles
\n0.00526 mole of Na2<\/sub>CO3<\/sub> requires HCl
\n= 0.00526 \u00d7 2 moles = 0.01052
\n1 mole of NaHCO3<\/sub> required HCl = 1 mole
\n0.00526 mole of NaHCO3<\/sub> required HCl
\n= 0.00526 mole
\n\u2234Total HCl required = 0.01052 + 0.00526
\n=0.01578 moles<\/p>\n

Step3: To calculate volume of 0.1M HCl
\n0.1 mole of 0.1 M HCl are present in 1000 mL of HCl
\n0.01578 mole of 0.1 M HCl will be present in HCl = \\(\\frac{1000}{0.1}\\) x 0.01578
\n= 157.8 ml.<\/p>\n

Question 7.
\nA solution is obtained by mixing 300 g of 25% solution and 400 g of 40% solution by mass. Calculate the mass percentage of the resulting solution.
\nAnswer:
\n300 g of 25% solutions contains solute = 75g
\n400 g of 40% solution contains solute = 160 g.
\nTotal solute = 160 + 75 = 235 g
\nTotal solution=300 + 400 = 700 g
\n% of solute in final solution = \\(\\frac{235}{700}\\) \u00d7 100 = 33.5%
\n% of water in the final solution = 100 – 33.5 = 66.5%<\/p>\n

\"KSEEB<\/p>\n

Question 8.
\nAn antifreeze solution is prepared from 222.6 g of ethylene glycol (C2<\/sub>H6<\/sub>O2<\/sub>) and 200 g of water. Calculate the molality of the solution. If the density of the solution is 1.072 g mL-1<\/sup>, then what shall be the molarity of the solution?
\nAnswer:
\nMass of the solute, C2<\/sub>H4<\/sub>(OH)2<\/sub> = 222.6g
\nMolar mass of C2<\/sub>H4<\/sub> (OH)2<\/sub> = 62 g mol-1<\/sup>
\n\"2nd
\nMass of the solvent = 200 g = 0.200 kg<\/p>\n

\"2nd
\nTotal mass of the solution = 422.6g
\nVolume of the solutions =<\/p>\n

\"2nd<\/p>\n

Question 9.
\nA sample of drinking water was found to be severely contaminated with chloroform (CHCl3<\/sub>) supposed to be a carcinogen. The level of contamination was 15 ppm (by mass):
\n(i) express this in percent by mass
\n(ii) determine the molality of chloroform in the water sample.
\nAnswer:
\n15 ppm means 15 parts in million (106<\/sup>) parts by mass in solution
\n\u2234 % by mass = \\(\\frac{15}{10^{6}} \\times\\) \u00d7 100=15 \u00d7 10-4<\/sup>
\nTaking 15 g chloroform in 106<\/sup>g
\nMolar mass of CHCl3<\/sub> = 12 + 1 + 3 \u00d7 35.5 = 119.5 g mol-1<\/sup>
\nMolality = \\(\\frac{15 \/ 119.5}{10^{6}}\\) \u00d7 1000= 1.25 \u00d7 10-4<\/sup> m<\/p>\n

Question 10.
\nWhat role does the molecular interaction play in a solution of alcohol and water?
\nAnswer:
\nThere is strong hydrogen bonding in alcohol molecules as well as water molecules. On mixing the molecular interactions are weekend. Hence, they show positive deviations from ideal behaviour. AS a result, the solution will have higher vapour pressure and lower boiling point than that of water and alcohol.<\/p>\n

Question 11.
\nWhy do gases always tend to be less soluble in liquids as the temperature is raised?
\nAnswer:
\nDissolution of gas in liquid in an exothermic process (Gas + solvent \u21cc solution + Heat). As the temperature is increased, equilibrium shifts backward.<\/p>\n

Question 12.
\nState Henry\u2019s law and mention some important applications?
\nAnswer:
\nHenry\u2019s law states that at constant temperature, the solubility of a gas in a liquid is directly proportional to the pressure of the gas.
\nApplications: It is used<\/p>\n

    \n
  • in the production of carbonated beverages<\/li>\n
  • in the deep sea diving<\/li>\n
  • in the function of lungs.<\/li>\n<\/ul>\n

    \"KSEEB<\/p>\n

    Question 13.
    \nThe partial pressure of ethane over a solution containing 6.56 x 10-3<\/sup> g of ethane is 1 bar. If the solution contains 5.00 x 10-2<\/sup> g of ethane, then what shall be the partial pressure of the gas?
    \nAnswer:
    \nApplying the relationship m = KH<\/sub> x p
    \nIn the first case, 6.56 x 102<\/sup> g bar-1<\/sup>
    \nIn the second case, 5.00 x 10-2<\/sup> g = (6.56 \u00d7 10-2<\/sup> gbar-1<\/sup>) \u00d7 p
    \n\"2nd<\/p>\n

    Question 14.
    \nWhat is meant by positive and negative deviations from Raoult\u2019s law and how is the sign of \u0394mix<\/sub>H related to positive and negative deviations from Raoult\u2019s law?
    \nAnswer:
    \nIn +ve deviation, A-B interactions are weaker than those between A-A or B-B. In such molecules, A or B will find it easier to escape than in pure state. This increases vapour .pressure. In case of -ve deviation, A-B interaction = A-A or B-B. This leads to decrease in vapour pressure.<\/p>\n

    In +ve deviation, \u0394mix<\/sub>H is + ve
    \nIn -ve deviation, \u0394mix<\/sub>H is – ve<\/p>\n

    Question 15.
    \nAn aqueous solution of 2% non-volatile solute exerts a pressure of 1.004 bar at the normal boiling point of the solvent. What is the molar mass of the solute?
    \nAnswer:
    \nVapour pressure of pure water at boiling
    \npoint (P\u00b0) = 1 atm = 1.013 bar
    \nvapour pressure of solution (ps<\/sub>) = 1.004 bar
    \nMass of solute = (w2<\/sub>) = 2g
    \nMass of solution = 100 g
    \nMass of solvent = 98g
    \nApplying Roault\u2019s law for dilute solution (being 2%)
    \n\"2nd<\/p>\n

    Question 16.
    \nHeptane and octane form an ideal solution. At 373 K, the vapour pressures of the two liquid components are 105.2 kPa and 46.8 kPa respectively. What will be the vapour pressure of a mixture of 26.0 g of heptane and 35 g of octane?
    \nAnswer:
    \nMolar mass of heptane (C7<\/sub> H16<\/sub>) = 100 g mol-1<\/sup>
    \nMolar mass of octane (C8<\/sub>H18<\/sub>) = 114 g mol-1<\/sup>
    \n\"2nd
    \nx (octane) = 1 – 0.456 = 0.544
    \np (heptane) = 0.456 x 105.2 kPa = 47.97 kPa
    \np (octane) = 0.544 x 46.8 kPa = 25.46 kPa
    \np total = 47.97 + 25.46 = 73.43 kPa<\/p>\n

    Question 17.
    \nThe vapour pressure of water is 12.3 kPa at 300 K. Calculate vapour pressure of 1 molal solution of a non-volatile solute in it.
    \nAnswer:
    \n1 molal solution means 1 mol of the solute in 1 kg of solvent (water)
    \n\"2nd
    \nor ps<\/sub> = 12.08 kPa<\/p>\n

    Question 18.
    \nCalculate the mass of a non-volatile solute (molar mass 40 g mol-1<\/sup>) which should be dissolved in 114 g octane to reduce its vapour pressure to 80%.
    \nAnswer:
    \nPs = 80% of p\u00b0 = 0.80 p\u00b0 solute = \\(\\frac{w}{40} r\\) mol solvent (octane) =
    \n\"2nd
    \n\"2nd<\/p>\n

    Question 19.
    \nA solution containing 30 g of non-volatile solute exactly in 90 g of water has a vapour pressure of 2.8 kPa at 298 K. Further, 18 g of water is then added to the solution and the new vapour pressure becomes 2.9 kPa at 298 K. Calculate:
    \n(i) molar mass of the solute
    \n(ii) vapour pressure of water at 298 K.
    \nAnswer:
    \nSuppose the molar mass of the solute = Mg mol-1<\/sup>
    \nn2<\/sub> (solute) = \\(\\frac{30}{M}\\) moles
    \n\"2nd
    \nAfter adding 18 g of water,
    \nn(H2<\/sub>O)i.e, n1<\/sub> = 6 moles
    \n\"2nd
    \n\"2nd
    \nDividing equation (i) by equation (ii), we get
    \n\"2nd<\/p>\n

    (ii) putting M = 23 in equation (i), we get
    \n\"2nd<\/p>\n

    Question 20.
    \nA 5% solution (by mass) of cane sugar in water has freezing point of 271 A. Calculate the freezing point of 5% glucose in water if freezing point of pure water is 273.15 K.
    \nAnswer:
    \nMolality of sugar solution =
    \n\"2nd
    \n\u2206Tf <\/sub>for sugar solution = 273.15 – 271 = 2.150<\/sup>
    \n\u2206Tf<\/sub> = Kf<\/sub> \u00d7 m \u2234 Kf<\/sub> = 2.15\/0.146
    \nMolality of glucose solution =
    \n\"2nd
    \nFreezing point of glucose solution = 273.15 – 4.09 = 269.06 k<\/p>\n<\/div>\n<\/div>\n<\/div>\n

    \"KSEEB<\/p>\n

    Question 21.
    \nTwo elements A and B form compounds having formula AB2<\/sub> and AB4<\/sub>. When dissolved in 20 g of benzene (C6<\/sub>H6<\/sub>), 1 g of AB2<\/sub> lowers the freezing point by 2.3 K whereas 1.0 g of AB4<\/sub> lowers it by 1.3 K. Tin molar depression – constant for benzene is 5.1 K kg mol-1<\/sup>. Calculate atomic masses of A and B.
    \nAnswer:
    \n\"2nd
    \n\"2nd
    \nSuppose atomic masses of A and B are \u2018a\u2019 and \u2018b\u2019 respectively. Then
    \nMolar mass of AB2<\/sub> = a + 2b = 110.87 g mol-1<\/sup>
    \nMolar mass of AB4<\/sub> = a + 4b =196.15 g mol-1<\/sup>
    \nEquation (ii) – Equation (i) gives
    \n2b = 85.28 orb = 42.64
    \nsubstituting in equation (i) we get
    \na + 2 \u00d7 42.64 = 110.87 or a = 25.59
    \nThus atomic mass A = 25.59 u
    \nAtomic mass of B = 4.64 u.<\/p>\n

    Question 22.
    \nAt 300 K, 36 g of glucose present in a litre of its solution has an osmotic pressure of 4.98 bar. If the osmotic pressure of the solution is 1.52 bars at the same temperature, what would be its concentration?
    \nAnswer:
    \n\u03c0 = CRT
    \n\u2234 In the first case,
    \n4.98 = \\(\\frac{36}{180}\\) \u00d7 R \u00d7 300 = 60R
    \nIn the second case, 1.52 = C \u00d7 R \u00d7 300
    \nDividing (ii) by (i), we get C = 0.061 M.<\/p>\n

    Question 23.
    \nSuggest the most important type of intermolecular attractive interaction in the following pairs.
    \n(i) n-hexane and n-octane
    \n(ii) I2<\/sub> and CCl4<\/sub>
    \n(iii) NaClO4<\/sub> and water
    \n(iv) methanol and acetone
    \n(v) acetonitrile (CH3<\/sub>CN) and acetone (C3<\/sub>H6<\/sub>O).
    \nAnswer:
    \n(i) Both are non-polar. Hence, inter molecular interactions in them will be London\/ dispersion forces (discussed in class XI)
    \n(ii) Both are non-polar. Hence, inter molecular interactions in them will be London\/ dispersion forces (discussed in class XI)
    \n(iii) NaClO4<\/sub> gives Na+ and ClO4<\/sub>–<\/sup> ions in the solution while water is polar molecule. Hence, intermolecular interactions in them will be ion – dipole interactions.
    \n(iv) Both are polar molecules. Hence intermolecular interactions in them will be dipole-dipole interactions.
    \n(v) Both are polar molecules. Hence intermolecular interactions in them will be dipole-dipole interactions.<\/p>\n

    Question 24.
    \nBased on solute-solvent interactions, arrange the following in order of increasing solubility in n*octane and explain. Cyclohexane, KCl, CH3<\/sub>OH, CH3<\/sub>CN.
    \nAnswer:
    \n(i) Cyclohexane and n-octane both are non-polar. Hence they mix completely in all proportions.
    \n(ii) KCl is an ionic compound while n-octane is nonpolar. Hence, KCl will not dissolve at all in n-octane.
    \n(iii) CH3<\/sub>OH and CH3<\/sub>CN both are polar but CH3<\/sub>CN is less polar than CH3<\/sub>OH. As the solvent is non-polar, CH3<\/sub>CN will dissolve more than CH3<\/sub>OH is n-octane.
    \nThus the order of solubility will be KCl< CH3<\/sub>OH < CH3<\/sub>CN < Cyclohexane.<\/p>\n

    Question 25.
    \nAmongst the following compounds, identify which are insoluble, partially soluble and highly soluble in water?
    \n(i) phenol
    \n(ii) toluene
    \n(iii) formic acid
    \n(iv) ethylene glycol
    \n(v) chloroform
    \n(vi) pentanol.
    \nAnswer:
    \n(i) Partially soluble (because phenol has
    \npolar-OH group but aromatic phenyl, C6<\/sub>H5 <\/sub>– group)
    \n(ii) Insoluble because toluene is non polar while water is polar
    \n(iii) Highly soluble because formic acid can form hydrogen bonds with water.
    \n(iv) Highly soluble because ethylene glycol can form hydrogen bonds with water
    \n(v) Insoluble chloroform is an organic liquid
    \n(vi) Partially soluble because-OH group is polar but the large hydrocarbon part (C5<\/sub>H11<\/sub>) is nonpolar.<\/p>\n

    \"KSEEB<\/p>\n

    Question 26.
    \nIf the density of some lake water is 1.25g mL-1<\/sup> and contains 92 g of Na+<\/sup> ions per kg of water, calculate the molality of Na+ ions in the lake.
    \nAnswer:
    \nNumber of moles in 92 g of Na+<\/sup> ions
    \n\"2nd
    \nAs these are present in 1 kg of water, by definition, molality = 4m.<\/p>\n

    Question 27
    \nIf the solubility product of CuS is 6 \u00d7 10-16<\/sup>, calculate the maximum molarity of CuS in aqueous solution.
    \nAnswer:
    \nMaximum molarity of CuS in aqueous solution = solubility of CuS in mol L-1<\/sup>
    \nIf S in the solubility of CuS in mol L-1<\/sup> , then
    \n\"2nd<\/p>\n

    Question 28.
    \nCalculate the mass percentage of aspirin (C9<\/sub>H8<\/sub>O4<\/sub>) in acetonitrile (CH3<\/sub>CN) when 6.5 g of C9<\/sub>H8<\/sub>O4<\/sub> is dissolved in 450 g of CH3<\/sub>CN.
    \nAnswer:
    \n\"2nd<\/p>\n

    Question 29.
    \nNalorphene (C19<\/sub>H21<\/sub>NO3<\/sub>), similar to morphine, is used to combat withdrawal symptoms in narcotic users. Dose of nalorphene generally given is 1.5 mg. Calculate the mass of 1.5 x 10-3<\/sup> m aqueous solution required for the above dose.
    \nAnswer:
    \n1.5 \u00d7 10-3<\/sup> m solution means that 1.5 \u00d7 10-3<\/sup> mole of nalorphene is dissolved in I kg of water.
    \nMolar mass of C19<\/sub>H21<\/sub>NO3<\/sub> = 19 \u00d7 12 + 21 + 14 + 48
    \n= 3119 mol-1<\/sup>
    \n\u2234 1.5 \u00d7 10-3<\/sup> mole of C19<\/sub>H21<\/sub>NO3<\/sub>;
    \n= 1.5 \u00d7 10-3<\/sup> \u00d7 3119 = 0.467 g = 467 mg
    \n\u2234 Mass of solution = 1000 g + 0.467 g \u2018
    \n=1000.467g
    \nThus, for 467 mg of nalorphene, solution required =1000.467 g for 1.5 mg of nalorphene, solution required =\\(\\frac{1000.467}{467} \\times 1.5 = 3.219\\)<\/p>\n

    Question 30.
    \nCalculate the amount of benzoic acid (C6<\/sub>H5<\/sub>COOH) required for preparing 250 mL of 0.15 M solution in methanol.
    \nAnswer:
    \n0.15 M solution means that 0.15 mole of benzoic acid is present in 1 L
    \ni.e. 1000 mL of the solution.
    \nMolar mass of benzoic acid (C6<\/sub>H5<\/sub>COOH) = 72 + 5 + 12 + 32 + 1 =122 g mol-1<\/sup>
    \n\u2234 0.15 mole of benzoic acid = 0.15 \u00d7 122g= 18.39
    \nThus, 1000 mL of the solution contain benzoic acid = 18.39
    \n\u2234 250 ml of the solution will contain benzoic acid = \\(\\frac{18.3}{1000}\\) \u00d7 250= 4.575g<\/p>\n

    \"KSEEB<\/p>\n

    Question 31.
    \nThe depression in freezing point of water observed for the same amount of acetic acid, trichloroacetic acid and trifluoroacetic acid increases in the order given above. Explain briefly.
    \nAnswer:
    \nThe depression in freezing points are in the order:
    \nacetic acid < trichloroacetic acid
    \n\"2nd
    \nFluorine, being most electronegative, has the highest electron withdrawing inductive effect. Consequently, trifluoroacetic acid is the Strongest acid while acetic acid is the weakest acid. Hence, trifluoroacetic acid ionizes to the largest extent while acetic acid ionizes to minimum extent to give ions.in their.solutions . in water. Greater the ions produced, greater is the depression in freezing point. Hence the depression in freezing point is maximum for fluoroacetic acid and minimum for acetic acid.<\/p>\n

    Question 32.
    \nCalculate the depression in the freezing point of water when 10 g of CH3<\/sub>CH2<\/sub> CHCICOOH is added to 250 g of water. Ka<\/sub> = 1.4 \u00d7 10-3<\/sup>, Kf<\/sub>= 1.86 K kg mol-1<\/sup>.
    \nAnswer:
    \nMolar mass of CH3<\/sub>CH2<\/sub>CHCICOOH = 15 + 14 + 13 + 35.5 + 45 = 122.5 g mol-1<\/sup>
    \n\"2nd
    \nIf \u03b1 is the degree of dissociation of CH3<\/sub>CH2<\/sub>CHCICOOH, then CH3<\/sub>CH2<\/sub>CHCLCOOH \u25a1 CH3CHXHCICOO–<\/sup> + H +
    \n\"2nd
    \nTo calculate vant’s Hoff factor:
    \nCH3<\/sub>CH2<\/sub>CHCICOOH \u25a1 CH3CH2<\/sub>HCICOO–<\/sup> + H+<\/sup>
    \n\"2nd<\/p>\n

    Question 33.
    \n19.5 g of CH2<\/sub>FCOOH is dissolved in 500 g of water. The depression in the freezing point of water observed is 1.0\u00b0C. Calculate the van\u2019t Hoff factor and dissociation constant of fluoroacetic acid. Kf<\/sub> of water is 1.86 K kg mol-1<\/sup>.
    \nAnswer:
    \nHere, w2<\/sub> = 19.5 g, w1<\/sub>, = 500g, Kf <\/sub>= 1.86 K kg mol-1<\/sup>, (ATf<\/sub>) obs = 1.0\u00b0
    \n\"2nd
    \nM2<\/sub> (Calculated) for CH2<\/sub> FCOOH = 14+19+45 = 78 g mol-1<\/sup>
    \nVant Hoff factor (i) =
    \n\"2nd
    \ncalculation of dissociation constant. Suppose degree of dissociation at the given concentration is a.
    \nThen
    \n\"2nd
    \nTaking volume of the solutions as 500ml,
    \n\"2nd<\/p>\n

    \"KSEEB<\/p>\n

    Question 34.
    \nVapour pressure of water at 293 K is 17.535 mm Hg. Calculate the vapour pressure of water at 293 K when 25 g of glucose is dissolved in 450 g of water.
    \nAnswer:
    \nHere, P\u00b0 = 17.535 mm, w2<\/sub> = 25g,
    \nw1<\/sub> = 450g
    \nFor solute (glucose, C6<\/sub>H12<\/sub>O6<\/sub>, M2<\/sub> = 180 g mol-1<\/sup>
    \nFor solvent (H2<\/sub>O), M1<\/sub> = 18g mol-1<\/sup>
    \nApplying Raoult\u2019s law,
    \n\"2nd
    \nsubstituting the given values, we get
    \n\"2nd
    \nsubstituting the given values, we get
    \n\"2nd<\/p>\n

    Question 35.
    \nHenry\u2019s law constant for the molality of methane in benzene at 298 K is 4.27 x 10s mm Hg. Calculate the solubility of methane in benzene at 298 K under 760 mm Hg.
    \nAnswer:
    \nHere, KH<\/sub> = 4.27 \u00d7 105<\/sup> mm
    \np = 760 mm
    \nApplying Henry\u2019s law
    \nP = KH<\/sub> x
    \n\"2nd
    \ni.e, mole fraction of methane in benzene = 1.78 x 10-3<\/sup><\/p>\n

    Question 36.
    \n100 g of liquid A (molar mass 140 g mol-1<\/sup>) was dissolved in 1000 g of liquid B (molar mass 180 g mol-1<\/sup>). The vapour pressure of pure liquid B was found to be 500 torr. Calculate the vapour pressure of pure liquid A and its vapour pressure in the solution if the total vapour pressure of the solution is 475 Torr.
    \nAnswer:
    \nNumber of moles of a liquid
    \n\"2nd
    \nNumber of moles of a liquid B (Solvent)
    \n\"2nd
    \nMole fraction of B in the solution (xA<\/sub>)
    \n\"2nd
    \nMole fraction of B in the solution (xB<\/sub>)
    \n= 1-0.114 = 0.886
    \nAlso, given p0<\/sup>B<\/sub> = 500 Torr
    \nApplying Raoult\u2019s law,
    \nPA<\/sub> = xA<\/sub>P\u00b0A<\/sub> = 0.114 \u00d7 P\u00b0A <\/sub>
    \nPB<\/sub>= xB<\/sub> p\u00b0B<\/sub> = 0.886 \u00d7 500 = 443 Torr
    \np Total = pA <\/sub>+ pB<\/sub>
    \n475 = 0.114 P\u00b0A<\/sub> +443 or
    \n\"2nd
    \nSubstituting this value in equation (i), we get pA <\/sub>= 0.114x 280.7 Torr = 32 Torr<\/p>\n

    Question 37.
    \nBenzene and toluene form ideal solution over the entire range of composition. The vapour pressures of pure benzene and toluene at 300 K are 50. 71mm Hg and 32.06 mm Hg respectively. Calculate the mole fraction of benzene in the vapour phase if 80 g of benzene is mixed with 100 g of toluene.
    \nAnswer:
    \nMolar mass of benzene (C6<\/sub>H6<\/sub> = 78 g mol-1<\/sup>
    \nMolar mass of toluene (C6<\/sub>H5<\/sub>CH3<\/sub>) = 92 g mol-1<\/sup>
    \n\u2234 Number of moles in 80 g of benzene
    \n\"2nd
    \nNumber of moles in 100 g of toluene
    \n\"2nd
    \nIn the solution, mole fraction of benzene
    \n\"2nd
    \n=0.486
    \nmole fraction of toluene = 1-0,486 = 0.514
    \np\u00b0 Benzene = 50.71 mm, p\u00b0 Toluene = 32.06mm
    \nApplying Raoult\u2019s law,
    \np Benzene= xTolucno<\/sub> \u00d7 p\u00b0Tolucne<\/sub> =0.514 \u00d7 32.06 mm = 16.48mm
    \nMole fraction of benzene in vapour phase
    \n\"2nd<\/p>\n

    Question 38.
    \nThe air is a mixture of a number of gases. The major components are oxygen and, nitrogen with approximate proportion of 20% is to 79% by volume at 298 K. The water is in equilibrium with air at a pressure of 10 atm. At 298 K if the Henry\u2019s law constants for oxygen and nitrogen at 298 K are 3.30 \u00d7 107<\/sup> mm and 6.51 \u00d7 107<\/sup> mm respectively, calculate the composition of these gases in water.
    \nAnswer:
    \nTotal pressure of air in equilibrium with water = 10 atm
    \nAs air contains 20% oxygen and 79% nitrogen by volumes
    \npartial pressure of oxygen :
    \n(P ) = \\(\\frac{20}{100}\\) x 10 atm = 2atm
    \n= 2 x 760 mm = 1520 mm
    \npartial pressure of nitrogen \\(P_{N_{2}}\\)
    \n= \\(\\frac{79}{100}\\) \u00d7 10 atm = 7.9 atm = 7.9 x 760 mm
    \n= 6004 mm<\/p>\n

    KH<\/sub>(O2<\/sub>) = 3.30 x 107<\/sup>mm,KH<\/sub> (N2<\/sub>) = 6.51 \u00d7 107<\/sup> mm
    \nApplying Henry\u2019s law
    \n\"2nd<\/p>\n

    Question 39.
    \nDetermine the amount of CaCl (i= 2.47) dissolved in 2.51itre of water such that its osmotic pressure is 0.75 atm at 27\u00b0 C.
    \nAnswer:
    \n\"2nd
    \nMolar mass of CaCl2<\/sub>= 40 + 2x 35.5 = 111 gmol-1<\/sup>
    \nAmount dissolved = 0.0308 xlllg = 3.42g<\/p>\n

    Question 40.
    \nDetermine the osmotic pressure of a solution prepared by dissolving 25 mg of K2<\/sub>SO4<\/sub>in 2 litre of water at 25\u00b0 C, assuming that it is completely dissociated.
    \nAnswer:
    \nK2<\/sub>SO4<\/sub> dissolved = 0.025g
    \nvolume of solution = 2L
    \nT = 250\u00b0C = 298 K
    \nMolar mass of K2<\/sub>SO4<\/sub> =
    \n2×39 + 32 + 4 \u00d7 16= 174gmol-1<\/sup>
    \nAs, K2<\/sub>SO4<\/sub> dissociates completely as K2<\/sub>SO4<\/sub> \u2192 2K+ SO4<\/sub>2-<\/sup>
    \ni.e., ions produced = 3 \u2234 i =3
    \n\"2nd<\/p>\n

    \"KSEEB<\/p>\n

    2nd PUC Chemistry Solutions Additional Questions and Answers<\/h3>\n

    Question 1.
    \nHow does molarity of a solution change with temperature?
    \nAnswer:
    \nMolarity decreases with increase in temperature because volume of solution increases with increase in temperature.<\/p>\n

    Question 2.
    \nState Raoult\u2019s law.
    \nAns:
    \nIt states that for a solution of volatile liquids its vapour pressure of each component is directly proportional to their mole fraction in the solution.
    \nThe relative lowering of vapour pressure is equal to mole fraction of solute in case of non volatile solute.<\/p>\n

    Question 3.
    \nWhy is liquid ammonia bottle first cooled in ice before opening it?
    \nAnswer:
    \nAt room temperature, the vapour pressure of liquid ammonia is very high. On cooling, vapour pressure decreases. Hence the liquid ammonia will not splash out.<\/p>\n

    Question 4.
    \nSodium chloride and calcium chloride are used clear show from the roads. Why?
    \nAnswer:
    \nSodium chloride depresses the freezing point of water to such an extent that it cannot freeze to form ice. Therefore, it melts off easily at the prevailing temperature.<\/p>\n

    Question 5.
    \nTwo liquids A and B boil at 145\u00b0C and 190\u00b0C
    \nrespectively. Which of them has a higher vapour pressure of 80\u00b0C? (CBSE 2006)
    \nAnswer:
    \nA being more volatile will have higher vapour pressure at 80\u00b0C.<\/p>\n

    \"KSEEB<\/p>\n

    Question 6.
    \nCalculate the molality of sulphuric acid solution in which the mole fraction is 0.85.
    \nAnswer:
    \n\"2nd<\/p>\n

    Question 7.
    \nThe molar freezing point depression constant for benzene is 4.90 kg mol’1. Selenium exists as polymer. When 3.26 g of selenium is dissolved in 226 g of benzene, the observed freezing point is 0.112\u00b0C lower for pure benzene. Decide the molecular formula of selenium (At Wt. of selenium is 78.8 g mol-1<\/sup>). (CBSE 2002)
    \nAnswer:
    \n\"2nd<\/p>\n

    Question 8.
    \nCCI4<\/sub>\u00a0and water are immiscible whereas ethanol and water are miscible in all proportions. Correlate this behaviour with molecular structure of these compounds. (CBSE 2003,2001)
    \nAnswer:
    \nCCI4<\/sub>\u00a0is a non-polar covalent compound.
    \nWater is a polar compound. CCI4<\/sub> can neither form H–<\/sup> bonds with water molecules nor can it break H–<\/sup> bonds between water molecules. Therefore, it is insoluble in water.<\/p>\n

    Ethanol is a polar compound and can form H’ bonds with water, which is a polar solvent, therefore it is miscible with water in all proportions.<\/p>\n

    Question 9.
    \nThe molarity of a solution of sulphuric acid is 1.35 M. Calculate its molarity (The density of the acid solution is 1.02 g cm3<\/sup> )
    \nAnswer:
    \nLet the solution be 1 litre or 1000 cm3<\/sup>
    \n\u2234 Number of moles of H2<\/sub>SO4<\/sub>= 1.35
    \nWt. of solution = 1000 x 1.02 = 1020 g
    \nWt. of sulphuric acid = 1.35 x 98= 132.3g :
    \nWt. of water = 1020 – 132.3 = 887.79
    \nMolality of H2<\/sub>SO4<\/sub>= ppp x 1000 = 1.52 m<\/p>\n

    From (i) M, = 110.82, from (ii) M2 = 196.15 AB4 – AB, = B2 196.15- 110.82 = B2 85.33 = B2 B = 42.665
    \nMolar mass of AB2 = Atomic mass of A+ x 2 atomic mass of B 110.82=Atomic mass of A+85.33 Atomic mass of A = 110.82 – 85.33 = 25.49 Atomic mass of A = 25.499 Atomic mass of B = 42.669<\/p>\n

    Question 10.
    \nTwo elements A and B form compounds having molecular formula AB2 and AB4. When dissolved 20 g of C6H6, 1 g of AB2 lowers the freezing point by 2.3 K, whereas 1.0 g of AB\u201e4 lower it by 1.3 K. The molar depression constant for benzene is 5.1 kg mol*1. Calculate the mass of A and B. (CBSE2004)
    \nAnswer:
    \nLet the molar mass of AB2<\/sub> and AB4<\/sub> be M1<\/sub>and M2<\/sub>
    \nThen, for AB2<\/sub>,
    \n\"2nd
    \nFrom (I) M1<\/sub> = 110.82, from (ii) M1<\/sub> = 196.15
    \nAB4<\/sub>– AB2<\/sub>= B2<\/sub>
    \n196.15 – 110.82 =B2<\/sub>
    \n85.33 = B2<\/sub>
    \nB = 42.665
    \nMolar mass of AB2 = Atomic mass of A+ x 2 atomic mass of B
    \n110.82 = Atomic mass of A+ 85.33
    \n\u2234 Atomic mass of A = 110.82-85.33 = 25.49
    \nAtomic mass of A = 25 .499
    \nAtomic mass of B = 42.669<\/p>\n

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

    You can Download Chapter 2 Solutions Questions and Answers, Notes, 2nd PUC Chemistry\u00a0Question Bank with Answers Karnataka State Board Solutions help you to revise complete Syllabus and score more marks in your examinations. Karnataka 2nd PUC Chemistry Question Bank Chapter 2 Solutions 2nd PUC Chemistry Solutions NCERT Textbook Questions and Answers Question 1 Define the …<\/p>\n","protected":false},"author":5,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[1],"tags":[],"jetpack_sharing_enabled":true,"jetpack_featured_media_url":"","_links":{"self":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts\/10159"}],"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=10159"}],"version-history":[{"count":0,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts\/10159\/revisions"}],"wp:attachment":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/media?parent=10159"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/categories?post=10159"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/tags?post=10159"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}