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**7. The vapour pressure of ethanol and methanol are 44.5 mm and 88.7 Hg respectively. An ideal solution is formed at the same temperature by mixing 60 g of ethanol with 40 g of methanol. Calculate the total vapour pressure of the solution and the mole fraction of methanol in the vapour. (1986 - 4 Marks)**

**Ans : **66.17 mm, 0.65

**Solution : **** TIPS/Formulae :**

P_{total }= p_{A} + p_{B}

Molecular weight of CH_{3}OH = 12 + 3 + 16 + 1 = 32

Molecular weight of C_{2}H_{5}OH= 24 + 5 + 16 + 1 = 46

According to Raoultâ€™s law

P_{total} = p_{1} + p_{2}

where P_{total} = Total vapour pressure of the solution

p_{1} = Partial vapour pressure of one component

p_{2} = Partial vapour pressure of other component

Again, p_{1} = Vapour pressure Ã— mole fraction

Similarly, p_{2} = Vapour pressure x mole fraction

Mole fraction of CH_{3}OH =

Mole fraction of ethanol =

**NOTE THIS STEP :** Thus now let us first calculate the partial vapour pressures, i.e., p_{1} and p_{2} of the two component.

Partial vapour pressure of CH_{3}OH(p_{1})= 88.7 Ã— 0.49 = 43.48 mm

Partial vapour pressure of C_{2}H_{5}OH(p_{2}) = 44.5 Ã— 0.51 = 22.69 mm

âˆ´ Total vapour pressure of the solution = 43.48 + 22.69 mm = 66.17 mm

Mole fraction of CH3OH in vapour =

**8. The vapour pressure of a dilute aqueous solution of glucose (C _{6}H_{12}O_{6}) is 750 mm of mercury at 373 K. Calculate (i) molality and (ii) mole fraction of the solution. (1989 - 3 Marks)**

**Ans : **0.7503 mol/kg, 0.9868

**Solution : **** TIPS/Formulae :**

x_{2} (solute) = 1 â€“ 0.9868 = 0.0132

= 0.7503 mol kg^{â€“1}

**9. The vapour pressure of pure benzene at a certain temperature is 640 mm Hg. A non-volatile non-electrolyte solid weighing 2.175 g is added to 39.0 g of benzene. The vapour pressure of the solution is 600 mm Hg. What is the molecular weight of the solid substance? (1990 - 3 Marks)**

**Ans : **65.25

**Solution : TIPS/Formulae :**

According to Raoultâ€™s law,

Here, pÂº = 640 mm

p = 600 mm

w= 2.175 g

= 39.0

m = ?

M = 78

Substituting the various values in the above equation for Roult's law :

**10. The degree of dissociation of calcium nitrate in a dilute aqueous solution, containing 7.0 g. of the salt per 100 gm of water at 100ÂºC is 70%. If the vapour pressure of water at 100ÂºC is 760 mm, calculate the vapour pressure of the solution. (1991 - 4 Marks)**

**Ans : **746.3 mm Hg

**Solution : **** TIPS/Formulae** : First find moles of Ca(NO3)2 and water.

Then use the expression to find vapour pressure of solution.

Let initially 1 mole of Ca(NO_{3})_{2} is taken

Degree of dissociation of Ca(NO_{3})_{2} = 70/100 = 0.7

Ionisation of Ca(NO_{3})_{2} can be represented as

âˆ´ Total number of moles in the solution at equilibrium = (1 â€“ 0.7) + 0.7 + 2 Ã— 0.7 = 2.4

No. of moles when the solution contains 1 gm of calcium nitrate instead of 1 mole of the salt

**11. Addition of 0.643 g of a compound to 50 ml. of benzene (density : 0.879 g/ml.) lowers the freezing point from 5.51ÂºC to 5.03ÂºC. If Kf for benzene is 5.12 K kg molâ€“1, calculate the molecular weight of the compound. (1992 - 2 Marks)**

**Ans : **156.056

**Solution : **

**TIPS/Formulae :**

Given Wt. of benzene (solvent),

W = Volume Ã— density = 50 Ã— 0.879 = 43.95 g

Wt. of compound (solute), w = 0.643 g

Mol. wt. of benzene, M = 78; Mol. wt. of solute, m = ?

Depression in freezing point, DTf = 5.51 â€“ 5.03 = 0.48

Molal freezing constant, K_{f} = 5.12

Now we know that,

**12. What weight of the non-volatile solute, urea (NH _{2} â€“ CO â€“ NH_{2}) needs to be dissolved in 100g of water, in order to decrease the vapour pressure of water by 25%? What will be the molality of the solution? (1993 - 3 Marks)**

**Ans : **18.52 m

**Solution : **

** TIPS/Formulae :**

Here, w and m are wt. and molecular wt. of solute, W and M are wt. and molecular weight of solvent

p = Pressure of solution; pÂº = Normal vapour pressure

Let the initial (normal) pressure (pÂº) = p

âˆ´ Pressure of solution =

m = 60, M = 18, W = 100 gm

**13. The molar volume of liquid benzene (density=0.877 g mL ^{â€“1}) increases by a factor of 2750 as it vaporises at 20Â°C and that of liquid toluene (density=0.867 g mL^{â€“1}) increases by a factor of 7720 at 20Â°C. A solution of benzene and toluene at 20Â°C has a vapour pressure of 46.0 Torr. Find the mole fraction of benzene in the vapour above the solution. (1996 - 3 Marks)**

**Solution : **

**TIPS/Formulae :**

Volume of 1 mole of liq. benzene =78/0.877

Volume of 1 mole of toluene = 92/0.867

In vapour phase,

At 20Â°C, for 1 mole of benzene,

Similarly for 1mole of toluene,

As we know that, PV = nRT

Total vapour-pressure = 46 torr = 46/760 = 0.060 atm

Thus, 0.060 = 0.098 X_{B} + 0.029 (1 â€“ X_{B})

â‡’ 0.060 = 0.098 X_{B} + 0.029 â€“ 0.029 X_{B}

â‡’ 0.031 = 0.069 X_{B}

X_{B} + X_{T} = 1

X_{T} = 1â€“ 0.45 = 0.55 (in liquid phase)

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