(C) Vapour pressure expected by Raoult's law is calculated as:$P_{\text{expected}} = P_A^o X_A + P_B^o X_B = 45 \times \frac{1}{1+2} + 36 \times \frac

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(C) Vapour pressure expected by Raoult's law is calculated as:$P_{\text{expected}} = P_A^o X_A + P_B^o X_B = 45 \times \frac{1}{1+2} + 36 \times \frac{2}{1+2}$$P_{\text{expected}} = 45 \times \frac{1}{3} + 36 \times \frac{2}{3} = 15 + 24 = 39 \, \text{torr}$.Since the observed vapour pressure $(42 \, \text{torr})$ is greater than the expected vapour pressure $(39 \, \text{torr})$,the solution exhibits positive deviation from Raoult's law.Solutions showing positive deviation form minimum boiling azeotropes at specific compositions.

Class 12 Chemistry Solutions Mix Examples of Solutions

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$1 \, \text{mole}$ of liquid $A$ and $2 \, \text{moles}$ of liquid $B$ make a solution having an observed vapour pressure of $42 \, \text{torr}$. The vapour pressures of pure $A$ and pure $B$ are $45 \, \text{torr}$ and $36 \, \text{torr}$ respectively. The described solution:

A
is an ideal solution
B
shows negative deviation
C
is a minimum boiling azeotrope
D
has a volume less than the sum of individual volumes of both components.

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An aqueous solution of a non-volatile solute boils at $100.17^{\circ} C$. The temperature at which this solution will freeze (in $^{\circ} C$) is
$K_{b}(H_2 O) = 0.512^{\circ} C \ kg \ mol^{-1}$,
$K_{f}(H_2 O) = 1.86^{\circ} C \ kg \ mol^{-1}$

MediumTS EAMCET 2023

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Which one of the following statements is false?

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On mixing urea,the boiling point of $H_{2}O$ changed to $100.5^{\circ}C$. Calculate the freezing point of the solution,if $K_{f}$ of water is $1.87 \ K \cdot kg \cdot mol^{-1}$ and $K_{b}$ of water is $0.52 \ K \cdot kg \cdot mol^{-1}$. (in $^{\circ}C$)

DifficultTS EAMCET 2021

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At $300 \ K$,an ideal solution is formed by mixing $460 \ g$ of toluene with $390 \ g$ of benzene. If the vapour pressure of pure toluene and pure benzene at $300 \ K$ are $32 \ mm$ and $40 \ mm$ respectively,the mole fraction of toluene in the vapour phase is:

EasyAP EAMCET 2019

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An aqueous solution of a solute $AB$ has a $b.p.$ of $101.08^\circ C$ ($AB$ is $100\%$ ionized at the boiling point of the solution) and freezes at $-1.80^\circ C$. Given $K_b / K_f = 0.3$,the solute $AB$:

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$1 \, \text{mole}$ of liquid $A$ and $2 \, \text{moles}$ of liquid $B$ make a solution having an observed vapour pressure of $42 \, \text{torr}$. The vapour pressures of pure $A$ and pure $B$ are $45 \, \text{torr}$ and $36 \, \text{torr}$ respectively. The described solution:

Similar Questions

An aqueous solution of a non-volatile solute boils at $100.17^{\circ} C$. The temperature at which this solution will freeze (in $^{\circ} C$) is$K_{b}(H_2 O) = 0.512^{\circ} C \ kg \ mol^{-1}$,$K_{f}(H_2 O) = 1.86^{\circ} C \ kg \ mol^{-1}$

Which one of the following statements is false?

On mixing urea,the boiling point of $H_{2}O$ changed to $100.5^{\circ}C$. Calculate the freezing point of the solution,if $K_{f}$ of water is $1.87 \ K \cdot kg \cdot mol^{-1}$ and $K_{b}$ of water is $0.52 \ K \cdot kg \cdot mol^{-1}$. (in $^{\circ}C$)

At $300 \ K$,an ideal solution is formed by mixing $460 \ g$ of toluene with $390 \ g$ of benzene. If the vapour pressure of pure toluene and pure benzene at $300 \ K$ are $32 \ mm$ and $40 \ mm$ respectively,the mole fraction of toluene in the vapour phase is:

An aqueous solution of a solute $AB$ has a $b.p.$ of $101.08^\circ C$ ($AB$ is $100\%$ ionized at the boiling point of the solution) and freezes at $-1.80^\circ C$. Given $K_b / K_f = 0.3$,the solute $AB$:

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An aqueous solution of a non-volatile solute boils at $100.17^{\circ} C$. The temperature at which this solution will freeze (in $^{\circ} C$) is
$K_{b}(H_2 O) = 0.512^{\circ} C \ kg \ mol^{-1}$,
$K_{f}(H_2 O) = 1.86^{\circ} C \ kg \ mol^{-1}$