Question

The addition of $1.0\text{g}$ of a compound to $10\text{g}$ of water increases the boiling point by $0.3{}^{\circ }\text{C}.$ The amount of compound needed to prepare a $500\text{mL}$ $0.1\text{M}$ solution is:
(Given: Assume negligible dissociation or association of the compound, boiling point constant $K_b$ of water $=0.513{\text{K kg mol}}^{-1}$)

• A. $0.855\text{g}$
• B. $17.1\text{g}$
• C. $8.55\text{g}$
• D. $85.5\text{g}$

Answer 1

The correct option is C $8.55\text{g}$

Elevation of boiling point, $\Delta T=m{K}_b$
where, $K_b$ is the molal elevation constant.

Here, $K_b=0.513{\text{K kg mol}}^{-1}$ and $\Delta T=0.3{}^{\circ }\text{C}$

From the equation, $0.3=\frac{\left(\frac{1.0}{M}\right)\times 1000}{10}\times 0.513\Rightarrow M=171g\text{(molecular weight of solute)}\therefore \text{Molarity}=\frac{\left(\frac{\text{weight of solute}}{M}\right)\times 1000}{\text{Volume of solution (in mL)}}\Rightarrow 0.1=\frac{\frac{\text{weight of solute}}{171}\times 1000}{500}\therefore \text{Weight of the solute}=8.55\text{g}$