Question

The mass of a non-volatile solute of molar mass $60$ $gmo{l}^{-1}$ that should be dissolved in $126$ g of water to reduce its vapour pressure to $99\%$ will be.

• A. $2.8$ g
• B. $5.6$ g
• C. $8.4$ g
• D. $4.2$ g

Answer 1

Answer: (D)

$4.2$ g

$\frac{P^o-P}{P^o}=\frac{W_2}{M_2}\frac{M_1}{W_1}$

Here $W_1$ and $M_1$ are the mass and molar mass of water and $W_2$ and $M_2$ are the mass and molar mass of non-volatile solute.

$\frac{P^o-P}{P^o}$ is the relative lowering of vapour pressure which is $100-99=1$ %

$\frac{1}{100}=\frac{W_2\times 18}{60\times 126}$

$W_2=\frac{60\times 126}{18\times 100}=4.2$

Hence, 4.2 g of non-volatile solute should be dissolved in 126 g of water to reduce its vapour pressure to 99%.