Question

The elevation in boiling point, when 13.44 g of freshly prepared $CuC{l}_2$ are added to one kilogram of water is:

[Some useful data, $K_b$ = 0.52 $kgKmo{l}^{-1}$ . Molecular weight of $CuC{l}_2=$ 134.4 gm]

• A. $0.05$
• B. $0.1$
• C. $0.16$
• D. $0.21$

Answer 1

Answer: (C)

$0.16$

Elevation of boiling point ( $\Delta T_b$ ) is the difference between the boiling points of the solution and the pure solvent:

$\Delta T_b=T_b-{T_b}^0$

Elevation of boiling point can be calculated as

$\Delta T_b=i.m.K_b$

Here, m is the concentration of the solute expressed in molality and $K_b$ is the molal boiling point elevation constant of the solvent.

$i$ is van't Hoff factor $=3$ for $Cu{Cl}_2$

No. of moles of $Cu{Cl}_2$ is $\frac{13.44}{134.4}=0.1$

Molality of $Cu{Cl}_2$ is $m=\frac{0.1}{1}=0.1$

The elevation of boiling point is $3\times 0.1\times 0.52=0.156\approx 0.16$

Hence, the correct option is $C$