Question

Glycerin $\left(C_3{H}_8{O}_3\right)$ is a non-volatile liquid. What is the vapor pressure of a solution made by adding $\mathbf{164}\mathbf{}\mathit{g}$ of glycerin to $\mathbf{338}\mathbf{}\mathit{m}\mathit{L}$ of ${\mathit{H}}_{\mathbf{2}}\mathit{O}$ at $\mathbf{39}\mathbf{.}{\mathbf{8}}^{\mathbf{0}}\mathit{C}$? The vapor pressure of pure water at $\mathbf{39}\mathbf{.}{\mathbf{8}}^{\mathbf{0}}\mathit{C}$ is $\mathbf{54}\mathbf{.}\mathbf{74}\mathbf{}\mathit{t}\mathit{o}\mathit{r}\mathit{r}$ and its density is $\mathbf{0}\mathbf{.}\mathbf{992}\mathbf{}\mathit{g}\mathbf{/}\mathit{c}{\mathit{m}}^{\mathbf{3}}$ .

Short answer

The vapor pressure of a solution made by adding $164g$ of glycerin to 338mL of $H_{2}O$ at 39.8⁰C is 50 mmHg(approx)

Step-by-step solution

Calculating mole fractions of each component

$\begin{array}{c}Molesofwater=\frac{338mL\times 0.992g.m/L}{\mathrm{18.01.}g.mo{l}^{-1}}=18.62mol\\ Molesofglycerol=\frac{164g}{\mathrm{92.09.}g.mo{l}^{-1}}=1.781mol\\ {\chi }_{water}=\frac{18.62mol}{18.62mol+1.781mol}=0.913\\ {\chi }_{glycerol}=\frac{1.781mol}{18.62mol+1.781mol}=0.0873\end{array}$

Calculating vapour pressure

$\begin{array}{c}{\chi }_{water}+{\chi }_{glycerol}=1\\ Vaporpressure={\chi }_{water}\times \mathrm{54.74.}mm.Hg\\ =0.913\times \mathrm{54.74.}mm.Hg\\ =49.97or50\left(approx\right)\end{array}$