(B) Let $A$ be benzene and $B$ be toluene. Given: $P_A^o = 119 \ torr$,$P_B^o = 37.0 \ torr$,$x_B = 0.50$. Since $x_A + x_B = 1$,$x_A = 0.50$. Accordi

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(B) Let $A$ be benzene and $B$ be toluene. Given: $P_A^o = 119 \ torr$,$P_B^o = 37.0 \ torr$,$x_B = 0.50$. Since $x_A + x_B = 1$,$x_A = 0.50$. According to Raoult's Law,the partial pressures are: $P_A = P_A^o x_A = 119 \times 0.50 = 59.5 \ torr$ and $P_B = P_B^o x_B = 37.0 \times 0.50 = 18.5 \ torr$. The total pressure $P_{total} = P_A + P_B = 59.5 + 18.5 = 78.0 \ torr$. In the vapour phase,the mole fraction of toluene $(y_B)$ is given by Dalton's Law: $y_B = \frac{P_B}{P_{total}} = \frac{18.5}{78.0} \approx 0.237$.

Class 12 Chemistry Solutions Lowering of vapour pressure

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Vapour pressure of pure benzene is $119 \ torr$ and that of toluene is $37.0 \ torr$ at the same temperature. The mole fraction of toluene in the vapour phase,which is in equilibrium with a solution of benzene and toluene having a mole fraction of toluene $0.50$,will be:

JEE MAIN 2013 All JEE MAIN

A
$0.137$
B
$0.237$
C
$0.435$
D
$0.205$

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Lowering of vapour pressure

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$1 \ mol$ each of the following solutes are taken in $5 \ mol$ of water:
$(i) \ NaCl$
$(ii) \ K_2SO_4$
$(iii) \ Na_3PO_4$
$(iv) \ \text{glucose}$
Assuming $100 \%$ ionization of the electrolytes,the relative decrease in vapor pressure will be in the order:

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The vapour pressure of pure benzene and toluene are $0.9 \ Bar$ and $0.85 \ Bar$ respectively. Calculate the vapour pressure of the solution prepared by dissolving $7.8 \ g$ of benzene in $180 \ g$ of toluene. (in $Bar$)

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At $300 \ K$,when a solute is added to a solvent,its vapour pressure over the mercury reduces from $50 \ mm$ to $45 \ mm$. The value of the mole fraction of the solute will be:

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The vapour pressure of water at $20\,^{\circ}C$ is $17.54\,mm$. When $20\,g$ of a non-ionic substance is dissolved in $100\,g$ of water,the vapour pressure is lowered by $0.30\,mm$. What is the molecular weight of the substance?

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The vapour pressure of pure water is $23 \text{ mmHg}$. The vapour pressure of an aqueous solution,which contains $10$ mass per cent of solute '$A$' having molecular weight $50$ is (in $\text{ atm}$)

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Vapour pressure of pure benzene is $119 \ torr$ and that of toluene is $37.0 \ torr$ at the same temperature. The mole fraction of toluene in the vapour phase,which is in equilibrium with a solution of benzene and toluene having a mole fraction of toluene $0.50$,will be:

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$1 \ mol$ each of the following solutes are taken in $5 \ mol$ of water:$(i) \ NaCl$$(ii) \ K_2SO_4$$(iii) \ Na_3PO_4$$(iv) \ \text{glucose}$Assuming $100 \%$ ionization of the electrolytes,the relative decrease in vapor pressure will be in the order:

The vapour pressure of pure benzene and toluene are $0.9 \ Bar$ and $0.85 \ Bar$ respectively. Calculate the vapour pressure of the solution prepared by dissolving $7.8 \ g$ of benzene in $180 \ g$ of toluene. (in $Bar$)

At $300 \ K$,when a solute is added to a solvent,its vapour pressure over the mercury reduces from $50 \ mm$ to $45 \ mm$. The value of the mole fraction of the solute will be:

The vapour pressure of water at $20\,^{\circ}C$ is $17.54\,mm$. When $20\,g$ of a non-ionic substance is dissolved in $100\,g$ of water,the vapour pressure is lowered by $0.30\,mm$. What is the molecular weight of the substance?

The vapour pressure of pure water is $23 \text{ mmHg}$. The vapour pressure of an aqueous solution,which contains $10$ mass per cent of solute '$A$' having molecular weight $50$ is (in $\text{ atm}$)

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$1 \ mol$ each of the following solutes are taken in $5 \ mol$ of water:
$(i) \ NaCl$
$(ii) \ K_2SO_4$
$(iii) \ Na_3PO_4$
$(iv) \ \text{glucose}$
Assuming $100 \%$ ionization of the electrolytes,the relative decrease in vapor pressure will be in the order: