The vapor pressure of water at 23,^{circ}C is | vedclass

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(B) Given: Vapor pressure of pure water $(P^{\circ})$ = $19.8\,mm\,Hg$. Mass of water $(w_A)$ = $178.2\,g$. Molar mass of water $(M_A)$ = $18\,g/mol$.

Vedclass · Accepted Answer

$19.602$

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(B) Given: Vapor pressure of pure water $(P^{\circ})$ = $19.8\,mm\,Hg$. Mass of water $(w_A)$ = $178.2\,g$. Molar mass of water $(M_A)$ = $18\,g/mol$. Moles of water $(n_A)$ = $\frac{178.2}{18} = 9.9\,mol$. Moles of glucose $(n_B)$ = $0.1\,mol$. According to Raoult's Law,the relative lowering of vapor pressure is given by: $\frac{P^{\circ} - P_s}{P^{\circ}} = \frac{n_B}{n_A + n_B}$. Substituting the values: $\frac{19.8 - P_s}{19.8} = \frac{0.1}{9.9 + 0.1} = \frac{0.1}{10} = 0.01$. $19.8 - P_s = 19.8 	imes 0.01 = 0.198$. $P_s = 19.8 - 0.198 = 19.602\,mm\,Hg$.

The vapor pressure of water at $23\,^{\circ}C$ is $19.8\,mm\,Hg$. What will be the vapor pressure (in $mm\,Hg$) of the resulting solution when $0.1\,mol$ of glucose is dissolved in $178.2\,g$ of water?

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