(D) Given: $P_{H_2O}^{\circ} = 25 \ torr$. Moles of urea $(n_{u})$ = $\frac{12 \ g}{60 \ g \ mol^{-1}} = 0.2 \ mol$. Moles of glucose $(n_{g})$ = $\fr

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(D) Given: $P_{H_2O}^{\circ} = 25 \ torr$. Moles of urea $(n_{u})$ = $\frac{12 \ g}{60 \ g \ mol^{-1}} = 0.2 \ mol$. Moles of glucose $(n_{g})$ = $\frac{36 \ g}{180 \ g \ mol^{-1}} = 0.2 \ mol$. Moles of water $(n_{H_2O})$ = $\frac{100 \ g}{18 \ g \ mol^{-1}} = 5.55 \ mol$. Mole fraction of water $(\chi_{H_2O})$ = $\frac{n_{H_2O}}{n_{H_2O} + n_{u} + n_{g}} = \frac{5.55}{5.55 + 0.2 + 0.2} = \frac{5.55}{5.95} \approx 0.9328$. Vapour pressure of solution $(P_{H_2O})$ = $\chi_{H_2O} \times P_{H_2O}^{\circ} = 0.9328 \times 25 \ torr = 23.32 \ torr$.

Class 12 Chemistry Solutions Lowering of vapour pressure

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At $298 \ K$,the vapour pressure of pure water is $25 \ torr$. The vapour pressure of water,when $12 \ g$ of urea (molar mass,$60 \ g \ mol^{-1}$) and $36 \ g$ of glucose (molar mass,$180 \ g \ mol^{-1}$) is dissolved in $100 \ g$ of water at the same temperature (in $torr$) is

AP EAMCET 2022 All AP EAMCET

A
$25.02$
B
$24.12$
C
$23.92$
D
$23.32$

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The vapor pressure of a mixture of $300 \ mL$ of ethyl alcohol and propyl alcohol is $290 \ mm$. The vapor pressure of propyl alcohol is $200 \ mm$. If the mole fraction of ethyl alcohol is $0.6$,what will be its vapor pressure (in $mm$) at the same temperature?

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The number of moles of chlorobenzene and bromobenzene are $0.1$ and $0.2$ respectively. The vapour pressure of pure chlorobenzene and pure bromobenzene are $0.350 \ bar$ and $0.500 \ bar$ respectively. Calculate the total vapour pressure of the solution prepared by mixing these two components.

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If $P_0$ and $P_s$ are the vapour pressures of the solvent and its solution respectively,and $N_1$ and $N_2$ are the mole fractions of the solvent and solute respectively,then which of the following is correct?

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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?

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The vapour pressure of water at $20\, ^oC$ is $17.5\, mm\, Hg$. If $18\, g$ of glucose $(C_6H_{12}O_6)$ is added to $178.2\, g$ of water at $20\, ^oC$,the vapour pressure of the resulting solution will be .......... $mm\, Hg$.

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At $298 \ K$,the vapour pressure of pure water is $25 \ torr$. The vapour pressure of water,when $12 \ g$ of urea (molar mass,$60 \ g \ mol^{-1}$) and $36 \ g$ of glucose (molar mass,$180 \ g \ mol^{-1}$) is dissolved in $100 \ g$ of water at the same temperature (in $torr$) is

Similar Questions

The vapor pressure of a mixture of $300 \ mL$ of ethyl alcohol and propyl alcohol is $290 \ mm$. The vapor pressure of propyl alcohol is $200 \ mm$. If the mole fraction of ethyl alcohol is $0.6$,what will be its vapor pressure (in $mm$) at the same temperature?

The number of moles of chlorobenzene and bromobenzene are $0.1$ and $0.2$ respectively. The vapour pressure of pure chlorobenzene and pure bromobenzene are $0.350 \ bar$ and $0.500 \ bar$ respectively. Calculate the total vapour pressure of the solution prepared by mixing these two components.

If $P_0$ and $P_s$ are the vapour pressures of the solvent and its solution respectively,and $N_1$ and $N_2$ are the mole fractions of the solvent and solute respectively,then which of the following is correct?

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?

The vapour pressure of water at $20\, ^oC$ is $17.5\, mm\, Hg$. If $18\, g$ of glucose $(C_6H_{12}O_6)$ is added to $178.2\, g$ of water at $20\, ^oC$,the vapour pressure of the resulting solution will be .......... $mm\, Hg$.

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