Question

Lowering of vapour pressure in 1 molal aqueous solution at ${100}^{0}C$ is :

• A. 13.44 mm Hg
• B. 14.12 mm Hg
• C. 31.2 mm Hg
• D. 35.2 mm Hg

Answer 1

Answer: (A)

13.44 mm Hg

As we know

$\frac{\Delta P}{P_{H_{2}O}^0}=X_{solute}$

and also $\frac{X_{solute}\times 1000}{(1-X_{solute})m_{solvent}}=m$

$\frac{X_{solute}\times 1000}{(1-X_{solute})18}=1$

$1018X_{solute}=18$

$X_{solute}=\frac{18}{1018}=0\cdot 0176$

So, $\frac{\Delta P}{P_{H_{2}O}^0}=0\cdot 0176$

$\therefore \Delta P=0\cdot 0176\times P_{H_{2}O}^0$

$\Delta P=0\cdot 0176\times 760mmkg$

$\Delta P=13\cdot 438mmkg$

Hence, the correct option is $\text{A}$