Question

To observe an elevation of boiling point of $0.05^?C$, the amount of solute (Mol. $Wt. = 100$) to be added to $100\, g$ of water $(K_b=0.5)$ is

• A. 2 g
• B. 0.5 g
• C. 1 g
• D. 0.75 g

Answer 1

Answer: (C)

1 g

Answer 2

Elevation of boiling point,
$\Delta T_{b}=\frac{w \times K_{b} \times 1000}{M \times W}$
(Here, $w$ and $W=$ weights of solute and solvent respectively,
$M=$ molecular weight of solute and
$K_{b}=$ constant $)$
On substituting values, we get
$0.05=\frac{w \times 0.5 \times 1000}{100 \times 100}$
or $w=\frac{0.05 \times 100 \times 100}{0.5 \times 1000}=1\, g$