(B) Let $A$ be toluene and $B$ be benzene.Given: $X_{A} = 0.50$,$X_{B} = 1 - 0.50 = 0.50$.$P_{A}^{0} = 37.0 \ torr$,$P_{B}^{0} = 119 \ torr$.Total vap

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(B) Let $A$ be toluene and $B$ be benzene.Given: $X_{A} = 0.50$,$X_{B} = 1 - 0.50 = 0.50$.$P_{A}^{0} = 37.0 \ torr$,$P_{B}^{0} = 119 \ torr$.Total vapour pressure $P_{total} = P_{A}^{0} X_{A} + P_{B}^{0} X_{B} = (37.0 \times 0.50) + (119 \times 0.50) = 18.5 + 59.5 = 78.0 \ torr$.According to Dalton's Law,the mole fraction of toluene in the vapour phase $(y_{A})$ is given by $y_{A} = \frac{P_{A}}{P_{total}}$.$P_{A} = P_{A}^{0} X_{A} = 37.0 \times 0.50 = 18.5 \ torr$.$y_{A} = \frac{18.5}{78.0} \approx 0.237$.

Class 12 Chemistry Solutions Lowering of vapour pressure

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

NEET 2017 All NEET

A
$0.462$
B
$0.237$
C
$0.506$
D
$0.325$

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The vapour pressure of two liquids $P$ and $Q$ are $80$ and $60 \, torr$,respectively. The total vapour pressure of the solution obtained by mixing $3 \, mole$ of $P$ and $2 \, mole$ of $Q$ would be .......... $torr$.

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Find the molar mass of solute when $2 \ g$ is dissolved in $60 \ g$ of benzene,and the relative lowering of vapour pressure is $0.06$. (Molar mass of benzene is $78 \ g \ mol^{-1}$)

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According to Raoult's law,the relative lowering of vapour pressure for a solution is equal to which of the following?

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Vapour pressure of solvent $A$ is $0.90 \text{ atm}$. When a non-volatile solute is added,the vapour pressure drops to $0.60 \text{ atm}$. What is the mole fraction of $A$ in the solution?

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What is the vapour pressure of a solution containing $1 \ mol$ of a non-volatile solute in $36 \ g$ of water (in $mm \ Hg$)? $(P_1^0 = 400 \ mm \ Hg)$

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

Similar Questions

The vapour pressure of two liquids $P$ and $Q$ are $80$ and $60 \, torr$,respectively. The total vapour pressure of the solution obtained by mixing $3 \, mole$ of $P$ and $2 \, mole$ of $Q$ would be .......... $torr$.

Find the molar mass of solute when $2 \ g$ is dissolved in $60 \ g$ of benzene,and the relative lowering of vapour pressure is $0.06$. (Molar mass of benzene is $78 \ g \ mol^{-1}$)

According to Raoult's law,the relative lowering of vapour pressure for a solution is equal to which of the following?

Vapour pressure of solvent $A$ is $0.90 \text{ atm}$. When a non-volatile solute is added,the vapour pressure drops to $0.60 \text{ atm}$. What is the mole fraction of $A$ in the solution?

What is the vapour pressure of a solution containing $1 \ mol$ of a non-volatile solute in $36 \ g$ of water (in $mm \ Hg$)? $(P_1^0 = 400 \ mm \ Hg)$

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