**1.** Vapour pressure of CHCl_{3 } and CH_{2}Cl_{2} at 298K are 200 mm Hg and 415mm Hg
respectively.

**(i)** Calulate the vapour pressure of the solution by mixing 25.5 g of
CHCl_{3 } and 40 g of CH_{2}Cl_{2} at 298K

**(ii) **mole fraction of each component in vapour phase.

(Molar mass CHCl_{3 } =118.5CH_{2}Cl_{2} = 85)

**2.** The vapour pressures of pure liquids A and B are 450 and 700 mm Hg respectively at
350K. Find the composition of liquid mixture if the total pressure is 600 mm Hg. Also
find the composition in vapour phase.

**3.** Calculate the mass of non volatile solute which should be dissolved in 114 g of octane
to reduce its vapour pressure to 80%
(Molar mass of solute=40, Molar mass of octane=114)

**4. **Benzene and naphthalene forms ideal solution at 300K. Vapour pressure of pure
benzene and naphthalene are 50.71 and 32.06 mm Hg respectively at 300K. Find the
mole fraction of each components in vapour phase if 80 g of benzene mixed with 100 g
naphthalene.(Molar mass enzene = 78, naphthalene = 128)

**5.** Benzene and toluene forms an ideal solution at 300K Vapour pressure of pure benzene
and pure toluene are 160 mm and 60 mm Hg respectively. Calculate the total pressure
of the solution under the following conditions :

**a)** mixing equal number of moles of benzene and toluene.

**b)** equal mass of benzene and toluene. Also find the composition of benzene and
toluene in vapour phase in each case.

**6.** Vapour pressure of solution containing benzene and toluene is (180x+120) mm Hg
where x is the mole fraction of toluene. Find the p0 benzene and p^{0} toluene respectively.

**7.** A solution containing 1 mole of X and 3 moles of Y gave a vapour pressure of 550 mm
Hg. The same solution containing 1 mole of X and 4 moles of Y gave a vapour pressure of 560 mm Hg. Find vapour pressure of pure X and pure Y.

**8.** Vapour pressure of dilute solution of glucose is 750 mm Hg at 373K. Find the mole
fraction and molality of glucose in solution.

**9.** Vapour pressure of water is 12.3 kPa at 300K. Calculate the vapour pressure of 1 molal solute in it.

**10. **A solution containing 30 g of non volatile solute in 90 g of water has a vapour
pressure of 2.8 kPa at 300K. Further 18 g of water added to the solution changed the
vapour pressure to 2.9kPa. Calculate the molar mass of the solute added and vapour
pressure of water at 300K.

**11.** Vapour pressure of pure benzene at 300K is 640 mm Hg. 2.175 g of non volatile
solute in 39 g of benzene gave a vapour pressure of 600 mm Hg. Find the molar mass
of solute added.(Molar mass of benzene=78)

**12.** A very small amount of solute in 60 ml of benzene gave a vapour pressure of 98.88
mm Hg. Vapour pressure of pure benzene is 100 mm Hg at this temperature. Find the
molality of the solution. If D Tf = 0.73K find Kf of benzene.

**13.** 34.2 g of sucrose and 36 g of glucose dissolved in 81 g of water at 300K find the vapour pressure of the solution. Vapour pressure of water at 300K= 30 mm Hg.
Molar mass of sucrose=342, glucose = 180.

**14.**Two elements A and B forms two compounds AB_{2} and AB_{4} respectively. When
dissolved in 20 g of benzene 1 g of AB_{2} lowers the freezing point by 2.3K while 1 g of
AB_{4} decreased the freezing point by 1.3K. Calculate the atomic masses of A and B.
Kf of benzene= 5.1 K Kg mole^{-1}

**15.** At 300K 36 g of glucose in 1 liter solution exerted an osmotic pressure of 4.98 bar.
What would be the concentration of the solution at 300K if it exerts a pressure of
1.52 bar.

**16.** 5% solution of sucrose (Molar mass =342) is isotonic with 0.877% solution of urea.
Determine the molar mass of urea.

**17.** Calculate the freezing point and boiling point of 1M solution of KCl. Density of the
solution = 1.04 g/ml Molar mass of KCl= 74.5 Kb = 0.52 K Kg mole^{-1} Kf = 1.86 K Kg mole^{-1}. Assume KCl undergoes 90% dissociation.

**18.** BaCl_{2} and KCl mixed in 1: 1 molal ratio showed a boiling point elevation of 2.6K. Determine the mass of each solute in 100 g water. Kb = 0.52 K Kg mole-1 Assume that both the solute undergoes 100% dissociation. Molar mass KCl = 74.5 BaCl_{2} = 208

**19. **Kf of benzene is 4.90 K Kg mole^{-1}. 3.26 g of Se in 226 gram of benzene showed a freezing point depression of 0.112^{0}C. Determine the molecular formula of Se. Atomic mass of Se= 78.8 benzene = 78

** 20.** 1.8 g of glucose in 100 ml is added to 34.2 g of sucrose in 100 ml. Find the osmotic pressure of the resulting solution. R= 0.0821 l atm mol^{-1} K^{-1}

**21.** 2 grams of benzoic acid (molar mass =122) in 25 g benzene gave ∆Tf = 1.62K Kf of benzene is 4.90 K Kg mole^{-1} Find and % association of benzoic acid if it exists as a dimmer.

**22.** 0.6 ml of acetic acid having density 1.06 g/ml is dissolved in 1 liter of water.∆Tf =0.02050C. Find the Van’t Hoff factor and dissociation constant of the acid. Molar mass of acetic acid = 60 Kf = 1.86 K Kg mole^{-1}

**23. **5% solution of sucrose has a freezing point of 71K.Calculate the freezing point of 5% glucose in water. Freezing point of water is 273.15K.

**24.** Which of the following solution in water will have

**a)** lowest freezing point

**b)** highest freezing point.

**c) **lowest boiling point

**d)** highest boiling point? 0.1M NaCl, 0.1 M BaCl_{2}, 0.1 M urea and 0.1 M Al_{2} (SO_{4})^{3}

**25.** Calculate the depression in freezing point of water when 10 g of 2-Chloro butanoic acid is added to 250 g of water. Ka = 1.4x 10^{-3} Kf = 1.86 K Kg mole^{-1} Molar mass of 2-Chloro butanoic acid =122.5

**26.** 19.5 g of Fluoro ethanoic acid (molar mass =78) is dissolved in 500 g of water. DTf = 1^{0}C .Calculate the Van’t Hoff factor and dissociation constant of the acid.

**27. **Determine the amount of CaCl_{2}(i = 2.47) dissolved in 2.5 litres of water such that itsosmotic pressure is 0.75 atm at 300K.

Submitted By : Mr. R. Srinivas Vasudevamurthy

Email: vasudm@yahoo.co.uk