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Calculate the Mass Percentage of Benzene (C6H6) and Carbon Tetrachloride (Ccl4) If 22 G of Benzene Is Dissolved in 122 G of Carbon Tetrachloride

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Calculate the Mass Percentage of Benzene (C6H6) and Carbon Tetrachloride (Ccl4) If 22 G of Benzene Is Dissolved in 122 G of Carbon Tetrachloride Class XII – NCERT – Chemistry Chapter 2 Solutions INTEX QUESTIONS Question 1: Calculate the mass percentage of benzene (C6H6) and carbon tetrachloride (CCl4) if 22 g of benzene is dissolved in 122 g of carbon tetrachloride. Solution 1: Mass of C H Mass percentage of C H 66 100% 66 Total mass of the solution Mass of C H 66 100% Mass of C6 H 6 Mass of CCl4 22 100% 22 122 = 15.28% Mass of CCl Mass percentage ofCCl 4 100% 4 Total mass of the solution Mass of CCl 4 100% Mass of C6 H64 Mass of CCl 122 100% 22 122 = 84.72% Alternatively, Mass percentage of CCl4 =(100-15.28) = 84.72% Question 2: Calculate the mole fraction of benzene in solution containing 30% by mass in carbon tetrachloride. Solution 2: 30% by mass of C6H6 in CCl4 30 g C6H6 in 100 g solution n Number of moles of C6H6 ( CH66) = 30/78 = 0.385 mol (molar mass of C6H6 = 78g) 2.Solutions Class XII – NCERT – Chemistry Chapter 2 Solutions No. of moles of n 70 CCl44 CCl 0.455 154 n CH x 66 CH66 nn C6 H 6 CCl4 0.385 0.385 0.458 0.385 0.455 0.84 X CCl4 = 1 – 0.458 = 0.542 Question 3: Calculate the molarity of each of the following solution: (a) 30 g of Co(NO3)2. 6H2O in 4.3 L of solution (b) 30 mL of 0.5 M H2SO4 diluted to 500 mL. Solution 3: -1 (a) Molar mass of Co(NO3)2.6H2O = 310.7 g mol no. of moles = 30/310.7 = 0.0966 Vol. of solution = 4.3 L Molarity = 0.0966/4.3 = 0.022 M (b) 1000 mL of 0.5M H2SO4 contain H2SO4 = 0.5 mole 30 mL of 0.5 M H2SO4 contain H2SO4 = 0.5/1000 × 30 = 0.015 mole Volume of solution = 500mL = 0.5L Molarity = 0.015/0.5 = 0.03M Question 4: Calculate the mass of urea (NH2CONH2) required in making 2.5 kg of 0.25 molal aqueous solution. Solution 4: 0.25 molal aqueous solution to urea means that Moles of urea = 0.25 mole -1 Mass of solvent (NH2CONH2) = 60 g mol 0.25 mole of urea = 0.25 × 60 = 15g Mass of solution = 1000 + 15g = 1.015 kg 1.015 kg of urea solution contains 15g of urea 2.5 kg of solution contains urea = 15/1.015 × 2.5 = 37 g. Question 5: 2.Solutions Class XII – NCERT – Chemistry Chapter 2 Solutions Calculate (a) molality (b) molarity and (c) mole fraction of KI if the density of 20% (mass/mass) aqueous KI is 1.202 g mL-1 Solution 5: 20% aq. KOH solution 20g of KI in 100g solution Mass of solvent = 100 – 20= 80 g no.of moles of Kl (i) Molality massof solvent (kg) 0.120 1.5 mol kg1 0.080 (ii) Density of solution = 1.202 g mL-1 100 Volume of solution 83.2 mL 1.202 0.0832 L 0.120 Molarity = 1.44 M 0.0832 (iii) No of moles of Kl = 0.120 80 n 4.44 HO2 18 0.120 x kl 0.120 4.44 0.120 0.0263 4.560 Question 6: H2S, a toxic gas with rotten egg like smell, is used for the qualitative analysis. If the solubility of H2S in water at STP is 0.195 m, calculate Henry’s law constant. Solution 6: Solubility of H2S gas = 0.195 m = 0.195 mole in 1 kg of solvent 1 kg of solvent = 1000g 1000 55.55 moles 18 0.195 X HS2 0.195 55.55 0.195 0.0035 55.745 Pressure at STP = 0.987 bar Applying Henry’s law 2.Solutions Class XII – NCERT – Chemistry Chapter 2 Solutions PKX HSHHS22 P 0.987 K HS2 282bar H X 0.0035 HS2 Question 7: 8 Heny’s law constant for CO2 in water is 1.67 × 10 Pa at 298 K. Calculate the quantity of CO2 in 500 mL of soda water when packed under 2.5 atm CO2 pressure at 298 K. Solution 7: KH = 1.67 × 108 Pa P = 2.5 atm = 2.5 × 101325 Pa CO2 p CO2 2.5 101325 3 X 1.517 10 CO2 8 kH 1.67 10 For 500 mL of soda water, water 500 mL 500 500g 27.78 moles 18 n 27.78 moles HO2 n CO2 27.78moles 27.78 n 42.14 103 mole CO2 = 42.14m mole = 42.14 × 10-3× 44g = 1.854 g Question 8: The vapour pressure of pure liquids A and B are 450 and 700 mm Hg respectively, at 350 K. Find out the composition of the liquid mixture if total vapour pressure is 600 mm Hg. Also find the composition of the vapour phase. Solution 8: PAB450 mm , P 700 mm , Ptotal 600 mm As Ptotal= PA + PB XPXPAA 1 AB PPPXBABA 600 700 (450 700)X A Or xA= 0.40 2.Solutions Class XII – NCERT – Chemistry Chapter 2 Solutions xXBA 1 1 1.40 0.60 PAAA x P 0.40 450 180mm PBBA x P 0.60 700 420mm Mole fraction of A in vapour phase P 180 A 0.30 PPAB180 420 And, mole fraction of B in vapour phase = 1 - 0.30 = 0.70 Question 9: Vapour pressure of pure water at 298 K is 23.8 mm Hg. 50 g of urea (NH2CONH2) is dissolved in 850 g of water. Calculate the vapour pressure of water for this solution and its relative lowering. Solution 9: P = 23.8 mm Hg -1 W2 = 50g, M2(urea) = 60g mol -1 w1 = 850g, M1 (water) = 18g mol To find: Ps and (P- Ps)/ P Solution: Applying Raoult’s law, P P n w/ M s 2 2 2 P n n w// M w M 1 2 1 1 2 2 PP 50/60 s P850/18 50/60 0.83 0.017 47.22 0.83 Putting P = 23.8 mm, we have 23.8 mm, we have 23.8 P s 0.017 Ps 23.8 - Ps = 0.017 Ps or 1.017 Ps = 23.8 or Ps = 23.4 mm Question 10: Boiling point of water at 750 mm Hg is 99.63C. How much sucrose is to be added to 500 g of water such that it boils at 100C. 2.Solutions Class XII – NCERT – Chemistry Chapter 2 Solutions Solution 10: Given Tb = 100 - 99.63 = 3.37 Mass of water w1 = 500g -1 Molar mass of water, M1 = 18g mol -1 Molar mass of sucrose, M2 = 342g mol To find Mass of sucrose, w2 = ? We know, Tb = Kb × m w2 100 Kb Mw21 M21 w Tb 342 500 3.37 w2 1000Kb 1000 0.52 w2 = 1108.2g Mass of solute, w2 = 1.11 kg Question 11: Calculate the mass of ascorbic acid (Vitamin C, C6H8O6) to be dissolved in 75 g of acetic acid to -1 lower its melting point by 1.5C. Kf = 3.9 kg mol Solution 11: Given: Tf = 1.5 Mass of CH3COOH, w1 =75g -1 M1 = 60g mol -1 M2(C6H8O) = 176g mol -1 Kf = 3.9 kg mol To find: w2 = ? 1000Kwf 2 Solution: Applying M2 wT1 f M w T Or w2 21 f 1000 K f 176 75 1.5 Wg5.077 2 1000 3.9 Question 12: Calculate the osmotic pressure in pascals exerted by solution prepared by dissolving 1.0 g of polymer of molar mass 185,000 in 450 mL of water at 37C. Solution 12: Given V = 450mL = 0.45L 2.Solutions Class XII – NCERT – Chemistry Chapter 2 Solutions T = 37C = 310K R = 8.314 kPaL k-1 mol-1 To find: Solution: Applying the formula, n CRT RT V 1.0g n 185,000g mol1 11 P 8.314 103 Pa LK 1 mol 1 310K 185,000 0.45 = 30.96 Pa NCERT Exercise Question 1: Define the term solution. How many types of solutions are formed? Write briefly about each type with an example. Solution 1: A Solution is a homogeneous mixture of two or more chemically non-reacting substances. Types of solution: There are nine types of solution. Types of Solution Examples Gaseous solutions: (a) Gas in gas: Air, mixture of O2 and N2, etc. (b) Liquid in gas: Water vapour (c) Solid in gas: Camphor vapours in N2 gas, smoke etc. Liquid solutions (a) Gas in liquid: CO2 dissolved in water (aerated water), and O2 dissolved in water, etc. (b) Liquid in liquid: Ethanol dissolved in water, etc. (c) Solid in liquid: Sugar dissolved in water, saline water, etc. Solid Solutions (a) Gas in solid: Solution of hydrogen in palladium (b) Liquid in solid:Amalgams,eg.Na-Hg. (c) Solid in solid: Gold ornaments (Cu/Ag with Au). Question 2: Give an example of a solid solution in which the solute is a gas. Solution 2: Solution of hydrogen in palladium and dissolved gases in minerals 2.Solutions Class XII – NCERT – Chemistry Chapter 2 Solutions Question 3: Define the following terms: (i) Mole fraction (ii) Molality (iii) Molarity (iv) Mass percentage Solution 3: (i) Mole Fraction: It is defined as the ratio of the number of moles of the solute to the total number of moles in the solution. If A is the number of moles of solute dissolved in B moles of solvent, then Mole fraction of solute. nA X A ....(1) nnAB nB Mole fraction of solvent X B ...(2) nnAB Adding the above two equations, we get nA nBAB n n XXAB 1 nABABAB n n n n n i.e., XA + XB = 1 XA = 1 - XB or XB = 1 - XA (ii) Molality: It is defined as the number of moles of a solute present in 1000g (1kg) of a solvent.
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