(C) According to Raoult's Law for non-volatile solutes: $\frac{P^o - P_s}{P^o} = \frac{n_2}{n_1 + n_2} \approx \frac{n_2}{n_1} = \frac{w_2 \times M_1}

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(C) According to Raoult's Law for non-volatile solutes: $\frac{P^o - P_s}{P^o} = \frac{n_2}{n_1 + n_2} \approx \frac{n_2}{n_1} = \frac{w_2 \times M_1}{M_2 \times w_1}$.Given: $P^o = 195 \ mm \ Hg$,$P_s = 192.5 \ mm \ Hg$,$w_2 = 1 \ g$,$w_1 = 100 \ g$,$M_1 = 58 \ g/mol$.Substituting the values: $\frac{195 - 192.5}{195} = \frac{1 \times 58}{M_2 \times 100}$.$\frac{2.5}{195} = \frac{58}{100 \times M_2}$.$M_2 = \frac{58 \times 195}{2.5 \times 100} = \frac{11310}{250} = 45.24 \ g/mol$.

Class 12 Chemistry Solutions Lowering of vapour pressure

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In an experiment,$1 \ g$ of a non-volatile solute was dissolved in $100 \ g$ of acetone (molar mass $= 58 \ g/mol$) at $298 \ K$. The vapour pressure of the solution was found to be $192.5 \ mm \ Hg$. The molecular weight of the solute is (vapour pressure of pure acetone $= 195 \ mm \ Hg$). (in $.24$)

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A
$25$
B
$35$
C
$45$
D
$55$

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Lowering of vapour pressure

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$160 \ g$ of non-volatile solute '$A$' is dissolved in $54 \ mL$ of water at $373 \ K$. What is the vapour pressure of the aqueous solution of '$A$' (in $Torr$)? (Given: molecular weight of '$A$' = $160 \ g \ mol^{-1}$)

MediumAP EAMCET 2020

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The partial vapour pressure of any volatile component of a solution is equal to the vapour pressure of the pure component multiplied by its mole fraction in the solution is called:

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At $300 \ K$,the vapor pressures of two pure liquids $A$ and $B$ are $150 \ mm \ Hg$ and $100 \ mm \ Hg$,respectively. If the mole fractions of $A$ and $B$ in the solution are equal,then the mole fraction of $B$ in the vapor phase at the same temperature is:

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Consider a binary solution of two volatile liquid components $1$ and $2$. $x_1$ and $y_1$ are the mole fractions of component $1$ in liquid and vapour phase,respectively. The slope and intercept of the linear plot of $\frac{1}{x_1}$ vs $\frac{1}{y_1}$ are given respectively as :

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At $80\,^{\circ}C$,the vapour pressure of pure liquid '$A$' is $520\, mm\, Hg$ and that of pure liquid '$B$' is $1000\, mm\, Hg$. If a mixture solution of '$A$' and '$B$' boils at $80\,^{\circ}C$ and $1\, atm$ pressure,then the amount of '$A$' in the mixture is ......... $mol\, \%$. $(1\, atm = 760\, mm\, Hg)$

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$160 \ g$ of non-volatile solute '$A$' is dissolved in $54 \ mL$ of water at $373 \ K$. What is the vapour pressure of the aqueous solution of '$A$' (in $Torr$)? (Given: molecular weight of '$A$' = $160 \ g \ mol^{-1}$)

The partial vapour pressure of any volatile component of a solution is equal to the vapour pressure of the pure component multiplied by its mole fraction in the solution is called:

At $300 \ K$,the vapor pressures of two pure liquids $A$ and $B$ are $150 \ mm \ Hg$ and $100 \ mm \ Hg$,respectively. If the mole fractions of $A$ and $B$ in the solution are equal,then the mole fraction of $B$ in the vapor phase at the same temperature is:

Consider a binary solution of two volatile liquid components $1$ and $2$. $x_1$ and $y_1$ are the mole fractions of component $1$ in liquid and vapour phase,respectively. The slope and intercept of the linear plot of $\frac{1}{x_1}$ vs $\frac{1}{y_1}$ are given respectively as :

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