(B) According to Raoult's Law for relative lowering of vapour pressure:$\frac{P^0 - P_s}{P_s} = \frac{n_2}{n_1} = \frac{w_2 \times M_1}{M_2 \times w_1

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(B) According to Raoult's Law for relative lowering of vapour pressure:$\frac{P^0 - P_s}{P_s} = \frac{n_2}{n_1} = \frac{w_2 \times M_1}{M_2 \times w_1}$Given:$P^0 = 640 \ mm \ Hg$,$P_s = 600 \ mm \ Hg$,$w_2 = 2.175 \ g$,$w_1 = 39.0 \ g$,$M_1 (\text{benzene, } C_6H_6) = 78 \ g/mol$Substituting the values:$\frac{640 - 600}{600} = \frac{2.175 \times 78}{M_2 \times 39.0}$$\frac{40}{600} = \frac{2.175 \times 2}{M_2}$$\frac{1}{15} = \frac{4.35}{M_2}$$M_2 = 4.35 \times 15 = 65.25 \ g/mol$

Class 12 Chemistry Solutions Lowering of vapour pressure

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The vapour pressure of pure benzene at a certain temperature is $640 \ mm \ Hg$. $A$ non-volatile non-electrolyte solid weighing $2.175 \ g$ is added to $39.0 \ g$ of benzene. The vapour pressure of the solution is $600 \ mm \ Hg$. What is the molar mass of the solid substance?

A
$6.96$
B
$65.25$
C
$63.8$
D
None of these

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Lowering of vapour pressure

If the vapor pressure of a solution formed by dissolving $1 \ mol$ of a non-volatile solute in $4 \ mol$ of a solvent is $24.0 \ kPa$,what will be the vapor pressure of the pure solvent in $kPa$?

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If $8 \; g$ of a non-electrolyte solute is dissolved in $114 \; g$ of $n$-octane to reduce its vapour pressure to $80 \%$,the molar mass (in $g \; mol^{-1}$) of the solute is.
Given that the molar mass of $n$-octane is $114 \; g \; mol^{-1}$.

MediumNEET 2020

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The vapour pressure of a solvent decreases by $10 \ mm$ of $Hg$ when a non-volatile solute is added to the solvent. The mole fraction of the solute in the solution is $0.2$. What should be the mole fraction of the solvent if the decrease in the vapour pressure is to be $20 \ mm$ of $Hg$?

MediumAIIMS 2015

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At $300 \ K$,the vapour pressures of two pure liquids,$A$ and $B$ are $100 \ mm \ Hg$ and $500 \ mm \ Hg$,respectively. If in a mixture of $A$ and $B$,the total vapour pressure is $300 \ mm \ Hg$,the mole fractions of $A$ in the liquid and in the vapour phase,respectively,are:

MediumKVPY 2011

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At $300 \ K$,when a solute is added to a solvent,its vapour pressure over the mercury reduces from $50 \ mm$ to $45 \ mm$. The value of the mole fraction of the solute will be:

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The vapour pressure of pure benzene at a certain temperature is $640 \ mm \ Hg$. $A$ non-volatile non-electrolyte solid weighing $2.175 \ g$ is added to $39.0 \ g$ of benzene. The vapour pressure of the solution is $600 \ mm \ Hg$. What is the molar mass of the solid substance?

Similar Questions

If the vapor pressure of a solution formed by dissolving $1 \ mol$ of a non-volatile solute in $4 \ mol$ of a solvent is $24.0 \ kPa$,what will be the vapor pressure of the pure solvent in $kPa$?

If $8 \; g$ of a non-electrolyte solute is dissolved in $114 \; g$ of $n$-octane to reduce its vapour pressure to $80 \%$,the molar mass (in $g \; mol^{-1}$) of the solute is.Given that the molar mass of $n$-octane is $114 \; g \; mol^{-1}$.

The vapour pressure of a solvent decreases by $10 \ mm$ of $Hg$ when a non-volatile solute is added to the solvent. The mole fraction of the solute in the solution is $0.2$. What should be the mole fraction of the solvent if the decrease in the vapour pressure is to be $20 \ mm$ of $Hg$?

At $300 \ K$,the vapour pressures of two pure liquids,$A$ and $B$ are $100 \ mm \ Hg$ and $500 \ mm \ Hg$,respectively. If in a mixture of $A$ and $B$,the total vapour pressure is $300 \ mm \ Hg$,the mole fractions of $A$ in the liquid and in the vapour phase,respectively,are:

At $300 \ K$,when a solute is added to a solvent,its vapour pressure over the mercury reduces from $50 \ mm$ to $45 \ mm$. The value of the mole fraction of the solute will be:

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If $8 \; g$ of a non-electrolyte solute is dissolved in $114 \; g$ of $n$-octane to reduce its vapour pressure to $80 \%$,the molar mass (in $g \; mol^{-1}$) of the solute is.
Given that the molar mass of $n$-octane is $114 \; g \; mol^{-1}$.