(A) According to Raoult's law,the lowering of vapour pressure is given by $\Delta P = P^{\circ} - P_s = P^{\circ} \times X_{solute}$.Given $\Delta P =

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(A) According to Raoult's law,the lowering of vapour pressure is given by $\Delta P = P^{\circ} - P_s = P^{\circ} \times X_{solute}$.Given $\Delta P = 20 \ mm \ Hg$ and $X_{solute} = 0.5$.So,$20 = P^{\circ} \times 0.5$,which gives $P^{\circ} = 40 \ mm \ Hg$.Now,we want the new lowering of vapour pressure $\Delta P' = 10 \ mm \ Hg$.Using $\Delta P' = P^{\circ} \times X'_{solute}$,we get $10 = 40 \times X'_{solute}$.Therefore,$X'_{solute} = \frac{10}{40} = 0.25$.The mole fraction of the solvent is $X_{solvent} = 1 - X'_{solute} = 1 - 0.25 = 0.75 = \frac{3}{4}$.

Class 12 Chemistry Solutions Lowering of vapour pressure

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The vapour pressure of a solvent decreases by $20 \ mm$ of $Hg$ when a non-volatile solute is added to the solvent. The mole fraction of the solute in the solution is $0.5$. What should be the mole fraction of the solvent for the decrease in the vapour pressure to be $10 \ mm$ of $Hg$?

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

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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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At a certain temperature,$5 \ g$ of a non-electrolyte solute is dissolved in $100 \ g$ of water. The vapour pressure of the resulting solution is $2985 \ N/m^2$,and the vapour pressure of pure water is $3000 \ N/m^2$. Calculate the molar mass of the solute.

Medium

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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:

Easy

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$A$ solution of liquids $A$ and $B$ is given. The mole fraction of $A$ in the vapor phase is $x_1$ and in the liquid phase is $x_2$. If $P_A^o$ and $P_B^o$ are the vapor pressures of pure $A$ and $B$ respectively,then the total vapor pressure $P_{total}$ will be:

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The mass of a non-volatile solute of molar mass $40 \ g \ mol^{-1}$ that should be dissolved in $114 \ g$ of octane to lower its vapour pressure by $20 \%$ is (in $g$)

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The vapour pressure of a solvent decreases by $20 \ mm$ of $Hg$ when a non-volatile solute is added to the solvent. The mole fraction of the solute in the solution is $0.5$. What should be the mole fraction of the solvent for the decrease in the vapour pressure to be $10 \ mm$ of $Hg$?

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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 $mm \ Hg$)? $(P_1^0 = 400 \ mm \ Hg)$

At a certain temperature,$5 \ g$ of a non-electrolyte solute is dissolved in $100 \ g$ of water. The vapour pressure of the resulting solution is $2985 \ N/m^2$,and the vapour pressure of pure water is $3000 \ N/m^2$. Calculate the molar mass of the solute.

$1 \ mol$ 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:

$A$ solution of liquids $A$ and $B$ is given. The mole fraction of $A$ in the vapor phase is $x_1$ and in the liquid phase is $x_2$. If $P_A^o$ and $P_B^o$ are the vapor pressures of pure $A$ and $B$ respectively,then the total vapor pressure $P_{total}$ will be:

The mass of a non-volatile solute of molar mass $40 \ g \ mol^{-1}$ that should be dissolved in $114 \ g$ of octane to lower its vapour pressure by $20 \%$ is (in $g$)

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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: