Question

$24\, g$ of glucose $\left( C _{6} H _{12} O _{6}\right)$ is added to $160.4 \,g$ 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 $=1 \,atm .$

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

Let vapour pressure of the solution $= p _{ s }$

$n_{\text {solute }}\left( C _{6} H _{12} O _{6}\right)=\frac{24}{180}=0.13 \,mol$

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

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

By Raoult's law, $\frac{p-p_{s}}{p^{\circ}}=\chi_{\text {solute }}$

$ \Rightarrow \frac{760- p _{ s }}{760}=\frac{0.13}{9.04}$

$9.04\left(760- p _{ s }\right)=760 \times 0.13$

$6870.4-9.04\, p _{ s }=98.8$

$6870.4-98.8=9.04\, p _{ s }$

$6771.6=9.04\, p _{ s }$

$p_{s}=\frac{6771.6}{9.04}$

$\therefore p _{ s }=749.07$