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151. Benzene and toluene form iedal solution over the entire range of composition. The vapour pressure of pure benzene and naphthalene at 300 K are 50.71 mm Hg and 32.06 mm Hg respectively. Calculate the mole-fraction of benzene in vapour phase if 80 g of benzene is mixed with 100 g of toluene.

Solution

Given that

P^o_Benzene= 50.51 mm Hg

P^o_Naphthalene= 50.51 mm Hg

Mass of Benzene = 80 g

Mass of Toluene = 100 g

Molar mass of benzene(C₆H₆) = 6 × 12 + 6 × 1 = 78 g mol ^{- 1}

Molar mass of toluene(C₆H₅CH₃) = 6 × 12 + 5 × 1 + 12 + 3 × 1 = 92 g mol^{– 1}

Use the formula

Mole of benzene $(n_{A})$
$= \frac{Mass of bezene}{Molar mass of bezene (C_{6} H_{6})} = \frac{80}{78 g {mol}^{- 1}} = 1.025 mol .$

Mass of Toluene $(n_{B})$
$= \frac{Mass of toluene}{Molar mass of toulene (C_{7} H_{8})} = 1 = \frac{100}{92 g {mol}^{- 1}} = 1.087 m o l$

Mole fraction of benzene $(x_{A})$
$= \frac{1.025 mol}{(1.025 mol + 1.087 mol)} = \frac{1.025}{2.113} = 0.486$

Similarly

Mole fraction of toluene, X _Toluene =

1-X_Benzene= 1 - 0.486 = 0.514

Use the formula if Henry law

P_A = p^o_A × X_A

Partial vapour pressure of benzene, P_Benzene= p^o_Benzene × XB_enzene

P_Benzene=0.487 × 50.71 = 24.645 mm Hg

Similarly

Partial pressure of Toluene, P toluene = 0.514 × 32.06 = 16.48 mmHg

Use the formula of mole fraction using partial pressure

Mole fraction of benzene = \frac{P_{benzene}}{P_{benzene} + P_{Toluene}}

Plug the values we get

Mole fraction of benzene = 24.645 /(24.645 + 16.48 )

= 24.645/41.123 = 0.60

156 Views

152. 100 g of liquid A (molar mass 140 g mol^–1) was dissolved in 1000 g of liquid B (molar mass 180 g mol^–1). The vapour pressure of pure liquid B and 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.

205 Views

153.

Heptane and Octane form an ideal solution at 373 K. The vapour pressures of the pure liquids at this temperature are 105.2 K Pa and 46.8 K Pa respectively. If the solution contains 25 g of heptane and 28.5 g of octane, calculate
(i) Vapour pressure exerted by heptane.
(ii) Vapour pressure exerted by solution.
(iii) Mole fraction of octane in the vapour phase.

370 Views

Short Answer Type

154. Ethylene glycol (molar mass = 62 g mol^–1) is a common automobile auto freeze. Calculate the freezing point of a solution containing 12.4 g of this substance in 100 g of water. Would it be advisable to keep the substance in the car radiator during summer?

159 Views

155. Calculate the temperature at which a solution containing 54 g of glucose, C₃H₁₂O₆, in 250 g of water will freeze. [K_ffor water = 1.86 K kg mol^–1]

472 Views

Long Answer Type

156. A solution containing 8 g of a substance in 100 g of diethyl ether boils at 36.86⁰C, whereas pure ether boils at 35.60⁰C. Determine the molecular mass of the solute. (For ether K_b= 2.02 K kg mol^–1) .

199 Views

157. How is osmotic pressure of a solution determined ? If the membrane used was slightly leaky, how will it influence the measured value of osmotic pressure?
Osmotic pressure of a solution containing 7 g of a protein per 103 ml of solution is 25 mm Hg at 310 K. Calculate the molecular mass or the protein. (R = 0.0821 L atm K^–1 mol).

164 Views

158. Define vapour pressure of a liquid. What happens to the vapour pressure when (a) a volatile solute dissolves in the liquid and (b) the dissolved solute is non-volatile?

174 Views

159. Discuss the various types of plots between the partial vapour pressure and the mole fractions of two components of the completely miscible liquids in a solution.

153 Views

Short Answer Type

160. Derive the relationship between relative lowering of vapour pressure and mole fraction of a volatiles liquid.

2177 Views