(C) Step $1$: Calculate the number of moles of solute $(n_2)$ and solvent $(n_1)$.$n_2 = \frac{38.4 \ g}{384 \ g \ mol^{-1}} = 0.1 \ mol$$n_1 = \frac{

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(C) Step $1$: Calculate the number of moles of solute $(n_2)$ and solvent $(n_1)$.$n_2 = \frac{38.4 \ g}{384 \ g \ mol^{-1}} = 0.1 \ mol$$n_1 = \frac{116 \ g}{58 \ g \ mol^{-1}} = 2 \ mol$Step $2$: Calculate the mole fraction of the solvent $(x_1)$.$x_1 = \frac{n_1}{n_1 + n_2} = \frac{2}{2 + 0.1} = \frac{2}{2.1} \approx 0.9524$Step $3$: Apply Raoult's law to find the vapour pressure of the solution $(P)$.$P = x_1 \times P_0 = 0.9524 \times 0.842 \ atm \approx 0.8019 \ atm$.Note: Based on the provided options,the closest value is $0.7999 \ atm$.

Class 12 Chemistry Solutions Lowering of vapour pressure

Medium

$38.4 \ g$ of unknown substance (molar mass $384 \ g \ mol^{-1}$) and $116 \ g$ of acetone is used to prepare a solution at $313 \ K$. If vapour pressure of pure acetone (molar mass $58 \ g \ mol^{-1}$) is $0.842 \ atm$,what is the vapour pressure of the solution (in $atm$)?

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A
$0.650$
B
$0.880$
C
$0.7999$
D
$0.958$

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Lowering of vapour pressure

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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?

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Calculate the relative lowering of vapour pressure if the vapour pressure of the pure solvent and the vapour pressure of the solution at $25^{\circ} C$ are $32 \ mm \ Hg$ and $30 \ mm \ Hg$ respectively.

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Calculate the mole fraction of pure liquid $B$ in solution if total vapour pressure of solution,vapour pressure of pure liquid $A$ and vapour pressure of pure liquid $B$ are $500 \ mmHg$,$400 \ mmHg$ and $575 \ mmHg$ respectively at given temperature.

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The vapour pressure of a pure solvent is $0.70 \ atm.$ When a non-volatile solute $A$ is dissolved in it,the vapour pressure of the solution becomes $0.35 \ atm.$ The mole fraction of solute $A$ will be$:-$

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An aqueous solution containing $28\%$ by mass of a liquid $A$ (molar mass $= 140 \ g/mol$) has a vapour pressure of $160 \ mm$ at $37 \ ^\circ C$. Find the vapour pressure of the pure liquid $A$ approximately .......... $mm$. (The vapour pressure of pure water at $37 \ ^\circ C$ is $150 \ mm$)

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$38.4 \ g$ of unknown substance (molar mass $384 \ g \ mol^{-1}$) and $116 \ g$ of acetone is used to prepare a solution at $313 \ K$. If vapour pressure of pure acetone (molar mass $58 \ g \ mol^{-1}$) is $0.842 \ atm$,what is the vapour pressure of the solution (in $atm$)?

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. If the vapour pressure of the solution is $600 \ mm$ of $Hg$,what is the molecular weight of the solid substance?

Calculate the relative lowering of vapour pressure if the vapour pressure of the pure solvent and the vapour pressure of the solution at $25^{\circ} C$ are $32 \ mm \ Hg$ and $30 \ mm \ Hg$ respectively.

Calculate the mole fraction of pure liquid $B$ in solution if total vapour pressure of solution,vapour pressure of pure liquid $A$ and vapour pressure of pure liquid $B$ are $500 \ mmHg$,$400 \ mmHg$ and $575 \ mmHg$ respectively at given temperature.

The vapour pressure of a pure solvent is $0.70 \ atm.$ When a non-volatile solute $A$ is dissolved in it,the vapour pressure of the solution becomes $0.35 \ atm.$ The mole fraction of solute $A$ will be$:-$

An aqueous solution containing $28\%$ by mass of a liquid $A$ (molar mass $= 140 \ g/mol$) has a vapour pressure of $160 \ mm$ at $37 \ ^\circ C$. Find the vapour pressure of the pure liquid $A$ approximately .......... $mm$. (The vapour pressure of pure water at $37 \ ^\circ C$ is $150 \ mm$)

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