At 298 K,the vapour pressure of pure water i | vedclass

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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 $(n_{g})$ = $\fr

Vedclass · Accepted Answer

$23.32$

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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 $(P_{H_2O})$ = $\chi_{H_2O} 	imes P_{H_2O}^{\circ} = 0.9328 	imes 25 \ torr = 23.32 \ torr$.

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

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