CLASS XII CHEMISTRY HOLIDAY HOME WORK 2016 chapter SOLUTIONS

SOLUTIONS

1.An antifreeze solution is prepared from 222.6 g of ethylene glycol C2H4(OH)2 and 200 g of water. calculate the molality of the solution. If density of this solution is 1.072 g/mL what will be the molarity of the solution?

2. A 0.1539 molal aqueous solution of cane sugar ( mol.mass =342 g/mol) has a freezing point of 271 K while the freezing point of pure water is 273.15 K. What will be the freezing point of an aqueous solution of 5 g gluose( mol.mass= 180 g/mol) per 100 g of solution?

3. The vapour pressure of water is 12.3 k.Pa at 300 K. Calculate the vapour pressure of a one molal solution of a non-volatile , non-ionic solute in water?

4. State Raoult's law for solutions of volatile liquids. Taking suitable examples explain the meaning of positive and negative deviation from Raoult's law?

5. Define the term osmotic pressure. Describe how molecular mass of a substance can be determined by a method based on measurement of osmotic pressure?

6. State the co-relating the osmotic pressure of a gas and its solubility in a liquid. State an application of this law?

7. Calculate the temperature at which a solution containing 54 gof glucose in 250 g of water will freeze.(kf for water = 1.86 K kg/mol).

8. A solution containing 8 g of a substance in 100 g diethyl ether boils at 36.86 degree celsius , where as pure ether boils at 35.6 degree celsius. Determine the molecular mass of the solute? (kb for ether = 2.02 Kkg/mol)?

9.(a) State how vapour pressure of a solvent is affected when a non-volatile solute is dissolved in it? (b) 5% solution (by mass) of cane sugar in water has freezing point of 271 K. Calculate the freezing point of 5% glucose (by mass) solution in water. Freezing point of pure water is 273.15 K. (molar mass o cane sugar =342 g/mol, glucose =180 g/mol)?

10.Calculate freezing point depression expected for 0.0711 m aqueous solution of Na2SO4. If this solution actually freezes at -0.320 degree celsius, what would be the value of van't Hoff factor?( kf = 1.86 C kg/mol).

11. 15 g of an unknown moleular substance was dissolved in 450 g of water. The resulting solution freezes at -0.34 degree celsius. What is the molar mass of the substance?( kf=1.86 K kg/mol).

12. 0.1 mole acetic acid was dissolved in 1 kg of benzene. Depression in freezing point of benzene was determined to be 0.256 K. What conclusion can you draw about the state of solute in solution? 9 kf of benzene = 5.12 K kg/mol).

13.define the following: mole fraction, van't Hoff factor, ideal solution, isotonic solutions, ebullioscopic constant.

14. A 0.561 m solution of unknown electrolyte depress the freezing point of water by 2.93 degree celsius. What is van't Hoff factor for this electrolyte. kf = 1.86 C kg/mol?

15. An aqueous solution containing 12.48 g barium chloride in 1 kg water bolis at 373.0832 K, calculate the degree of dissociation of barium chloride in water, molar mass of it =208.34 g/mol, kb= 0.52 Kkg/mol?