Question

$24g$ of glucose $\left(C_6{H}_{12}{O}_6\right)$ is added to $160.4g$ of water. The vapour pressure of water for this aqueous solution at $100^{\circ }C$ is

• A. 759.00 torr
• B. 749.07 torr
• C. 76.00 torr
• D. 752.40 torr

Answer 1

Answer: (B)

At $100^{\circ }C$, vapour pressure of water $=1atm.$

Thus, $p^{\circ }=760$ torr

Let vapour pressure of the solution $=p_s$

$n_{\text{solute }}\left(C_6{H}_{12}{O}_6\right)=\frac{24}{180}=0.13mol$

solvent $\left(H_{2}O\right)=\frac{160.4}{18}=8.91mol$

${\chi }_{\text{solute }}=\frac{0.13}{8.91+0.13}=\frac{0.13}{9.04}$

By Raoult's law, $\frac{p-p_s}{p^{\circ }}={\chi }_{\text{solute }}$

$\Rightarrow \frac{760-p_s}{760}=\frac{0.13}{9.04}$

$9.04\left(760-p_s\right)=760\times 0.13$

$6870.4-9.04p_s=98.8$

$6870.4-98.8=9.04p_s$

$6771.6=9.04p_s$

$p_s=\frac{6771.6}{9.04}$

$\therefore p_s=749.07$