Question

$18$ g glucose ($C_6{H}_{12}{O}_6$) is added to $178.2$ g of water. The vapour pressure of this aqueous solution at ${100}^{0}C$ in torr is:

• A. $7.60$
• B. $76.00$
• C. $752.40$
• D. $759.00$

Answer 1

Answer: (C)

$752.40$

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

For $178.2$g water, $n_A=\frac{178.2}{18}=9.9mol$

Molecular mass of glucose $=\left(6\right)\left(12\right)+\left(12\right)\left(1\right)+6\left(16\right)=180$g.

For $18$g glucose, $n_B=\frac{18}{180}=0.1mol$

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

$X_A=1-0.01=0.99$

For lowering of vapour pressure,

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

$P=760(1-0.01)$

$P=760-7.6$

$P=752.40$ torr

Vapour pressure of water is $752.40$ torr.

Therefore, the correct option is $C$.