Question

To observe an elevation of boiling point of $0.05^{°}C$, the amount of solute (Mol. $Wt.=100$) to be added to $100g$ 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)

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 $w=\frac{0.05\times 100\times 100}{0.5\times 1000}=1g$