Question

$18.0g$ of glucose$\left(C_6{H}_{12}{O}_6\right)$ is added to $178.2g$ of water. The vapour pressure of water for this aqueous solution at ${100}^{o}C$ is :

• A. $\mathrm{759..00}$ Torr
• B. $7.60$ Torr
• C. $76.00$ Torr
• D. $752.40$ Torr

Answer 1

The correct option is D $752.40$ Torr

Molecular mass of water $=2\times 1+1\times 16=18$g

For $178.2g$ water $n_A$=9.9

Molecular mass of glucose $=6\times 12+12\times 1+6\times 16=180$g

For 18g glucose $n_B$=0.1

$X_B=\frac{0.10}{(0.1+9.9)}=0.01$

$X_A=0.99$

For lowering of vapour pressure ,

$p=p_A^0{X}^A=P_A^0(1-X_B)$

$P=760(1-0.01)$

$=760-7.6$

$=752.4torr$