Home
Class 12
CHEMISTRY
Benzene and naphthalene form ideal solut...

Benzene and naphthalene form ideal 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 naphthalene.

A

0.0675

B

0.675

C

0.35

D

0.5

Text Solution

AI Generated Solution

The correct Answer is:
To solve the problem of finding the mole fraction of benzene in the vapor phase when 80 g of benzene is mixed with 100 g of naphthalene, we will follow these steps: ### Step 1: Calculate the number of moles of benzene and naphthalene 1. **Molar mass of benzene (C6H6)**: 78 g/mol 2. **Molar mass of naphthalene (C10H8)**: 128 g/mol **Number of moles of benzene (nA)**: \[ n_A = \frac{\text{mass of benzene}}{\text{molar mass of benzene}} = \frac{80 \, \text{g}}{78 \, \text{g/mol}} \approx 1.0256 \, \text{mol} \] **Number of moles of naphthalene (nB)**: \[ n_B = \frac{\text{mass of naphthalene}}{\text{molar mass of naphthalene}} = \frac{100 \, \text{g}}{128 \, \text{g/mol}} \approx 0.7813 \, \text{mol} \] ### Step 2: Calculate the mole fractions of benzene and naphthalene **Total moles (n_total)**: \[ n_{\text{total}} = n_A + n_B = 1.0256 + 0.7813 \approx 1.8069 \, \text{mol} \] **Mole fraction of benzene (X_A)**: \[ X_A = \frac{n_A}{n_{\text{total}}} = \frac{1.0256}{1.8069} \approx 0.567 \] **Mole fraction of naphthalene (X_B)**: \[ X_B = 1 - X_A = 1 - 0.567 \approx 0.433 \] ### Step 3: Calculate the partial pressures using Raoult's Law **Vapor pressure of pure benzene (P°_A)**: 50.71 mmHg **Vapor pressure of pure naphthalene (P°_B)**: 32.06 mmHg **Partial pressure of benzene (P_A)**: \[ P_A = X_A \cdot P°_A = 0.567 \cdot 50.71 \approx 28.75 \, \text{mmHg} \] **Partial pressure of naphthalene (P_B)**: \[ P_B = X_B \cdot P°_B = 0.433 \cdot 32.06 \approx 13.88 \, \text{mmHg} \] ### Step 4: Calculate the total pressure (P_total) \[ P_{\text{total}} = P_A + P_B = 28.75 + 13.88 \approx 42.63 \, \text{mmHg} \] ### Step 5: Calculate the mole fraction of benzene in the vapor phase Using Dalton's Law: \[ y_A = \frac{P_A}{P_{\text{total}}} = \frac{28.75}{42.63} \approx 0.6744 \] ### Final Answer The mole fraction of benzene in the vapor phase is approximately **0.6744**. ---
Promotional Banner

Similar Questions

Explore conceptually related problems

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

Equal moles of benzene and toluene are mixed. The vapour pressure of benzene and toluene in pure state are 700 and 600 mm Hg respectively. The mole fraction of benzene in vapour state is :-

The vapour pressure of pure benzene and toluene are 160 and 60 torr respectively. The mole fraction of toluene in vapour phase in contact with equimolar solution of benzene and toluene is:

A solution has 1:4 mole ratio of pentane to hexane . The vapour pressure of pure hydrocarbons at 20^@C are 440 mm Hg for pentane and 120 mm Hg for hexane .The mole fraction of pentane in the vapour phase is

Benzene and toulene form an ideal solution. 3 mole benzene and 2 mole toulene are added. V.P. of pure benezene and toulene are 300 & 200 mm of Hg respectively. The V.P of the solution (in mm of Hg) is

Two liquids A and B form an ideal solution of 1 mole of A and x moles of B is 550 mm of Hg. If the vapour pressures of pure A and B are 400 mm of Hg nd 600 mm of Hg respectively. Then x is-

Vapour pressures of para xylene and dimethylbenezene at 90^@C are respectivley 150 mm and 400 mm. Calculate the mole fraction of dimethyl benzene in the mixture that boils at 90^@C , when the pressure is 0.5 atm.

At 40^(@)C the vapour pressure of pure liquids, benzene and toluene, are 160 mm Hg and 60 mm Hg respectively. At the same temperature, the vapour pressure of an equimolar solution of the liquids, assuming the ideal solution will be:

The vapour pressure of pure liquid A and liquid B at 350 K are 440 mm and 720 mm of Hg. If total vapour pressure of solution is 580 mm of Hg then the mole fraction of liquid A in vapour phase will be :-

In a mixture of A and B having vapour pressure of pure A and B are 400 m Hg and 600 mm Hg respectively. Mole fraction of B in liquid phase is 0.5. Calculate total vapour pressure and mole fraction of A and B in vapour phase .