Question

The vapour pressure of $100$ g water reduces from $3000N{m}^{-2}$ to $2985N{m}^{-2}$ when $5g$ of substance $X$ is dissolved in it. The substance $X$ is :

• A. methanol
• B. glucose
• C. carbon dioxide
• D. none of the above

Answer 1

Answer: (B)

glucose

Raoult's law states that relative lowering of vapour pressure and is equal to

the mole fraction of the solute.

$\frac{p_1^0-p_1}{p_1}=x_2$

$p_1^0=3000N{m}^{-2}=$ vapour pressure of solution.

$p_1=2985N{m}^{-2}=$ vapour pressure of sollvent.

$x_2=$ mole-fraction of solute.

Let molecular wt. of $X$ is $x$.

No of moles of $X$ is $\frac{5}{x}$

No of moles of solvent is $\frac{100}{18}$.

Mole-fraction of $X$ is $\frac{\left(5/x\right)}{\left(100/18\right)}=9/10x$

So,

$\frac{3000-2985}{3000}=\frac{9}{10}x$

$⟹x=180$

The molecular weight of glucose is 180 g/mol.

Therefore, $X$ is glucose.