(A) Given: $\frac{P_A^o}{P_B^o} = \frac{1}{3}$ and $\frac{y_A}{y_B} = \frac{4}{3}$.According to Raoult's law and Dalton's law:$\frac{y_A}{y_B} = \frac

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(A) Given: $\frac{P_A^o}{P_B^o} = \frac{1}{3}$ and $\frac{y_A}{y_B} = \frac{4}{3}$.According to Raoult's law and Dalton's law:$\frac{y_A}{y_B} = \frac{P_A}{P_B} = \frac{P_A^o \cdot x_A}{P_B^o \cdot x_B}$Substituting the values:$\frac{4}{3} = \frac{1}{3} \cdot \frac{x_A}{x_B}$$\frac{x_A}{x_B} = 4 \Rightarrow x_A = 4x_B$Since $x_A + x_B = 1$,$4x_B + x_B = 1$ $\Rightarrow 5x_B = 1$ $\Rightarrow x_B = \frac{1}{5}$.

Class 12 Chemistry Solutions Lowering of vapour pressure

Difficult

Two liquids $A$ and $B$ have vapour pressure in the ratio $P_A^o : P_B^o = 1 : 3$ at a certain temperature. If the ratio of mole fractions of $A$ to $B$ in the vapour phase is $4 : 3$,then the mole fraction of $B$ in the solution at the same temperature is

A
$\frac{1}{5}$
B
$\frac{2}{3}$
C
$\frac{4}{5}$
D
$\frac{1}{4}$

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Lowering of vapour pressure

The vapour pressure of benzene at a certain temperature is $640 \, mm$ of $Hg$. $A$ non-volatile and non-electrolyte solid weighing $2.175 \, g$ is added to $39.08 \, g$ of benzene. The vapour pressure of the solution is $600 \, mm$ of $Hg$. What is the molecular weight of the solid substance?

DifficultAIPMT 1999

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Which of the following sets of variables gives a straight line with a negative slope when plotted? ($P =$ vapour pressure, $T =$ Temperature in $K$)

EasyTS EAMCET 2006

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Calculate the mass of a non-volatile solute (molar mass $40 \, g \, mol^{-1}$) which should be dissolved in $114 \, g$ octane to reduce its vapour pressure to $80 \%$.

Easy

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The vapour pressure of benzene at a certain temperature is $640 \ mm$ of $Hg$. $A$ non-volatile and non-electrolyte solid weighing $2.175 \ g$ is added to $39.08 \ g$ of benzene. If the vapour pressure of the solution is $600 \ mm$ of $Hg$,what is the molecular weight of the solid substance?

DifficultMHT CET 2008

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When $0.25 \text{ moles}$ of a non-volatile, non-ionizable solute was dissolved in $1 \text{ mole}$ of a solvent, the vapor pressure of the solution was $x\%$ of the vapor pressure of the pure solvent. What is $x$ (in $\%$)?

DifficultJEE MAIN 2026

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Two liquids $A$ and $B$ have vapour pressure in the ratio $P_A^o : P_B^o = 1 : 3$ at a certain temperature. If the ratio of mole fractions of $A$ to $B$ in the vapour phase is $4 : 3$,then the mole fraction of $B$ in the solution at the same temperature is

Similar Questions

The vapour pressure of benzene at a certain temperature is $640 \, mm$ of $Hg$. $A$ non-volatile and non-electrolyte solid weighing $2.175 \, g$ is added to $39.08 \, g$ of benzene. The vapour pressure of the solution is $600 \, mm$ of $Hg$. What is the molecular weight of the solid substance?

Which of the following sets of variables gives a straight line with a negative slope when plotted? ($P =$ vapour pressure, $T =$ Temperature in $K$)

Calculate the mass of a non-volatile solute (molar mass $40 \, g \, mol^{-1}$) which should be dissolved in $114 \, g$ octane to reduce its vapour pressure to $80 \%$.

The vapour pressure of benzene at a certain temperature is $640 \ mm$ of $Hg$. $A$ non-volatile and non-electrolyte solid weighing $2.175 \ g$ is added to $39.08 \ g$ of benzene. If the vapour pressure of the solution is $600 \ mm$ of $Hg$,what is the molecular weight of the solid substance?

When $0.25 \text{ moles}$ of a non-volatile, non-ionizable solute was dissolved in $1 \text{ mole}$ of a solvent, the vapor pressure of the solution was $x\%$ of the vapor pressure of the pure solvent. What is $x$ (in $\%$)?

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