Question

The molecular weight of an organic compound is 58.0 g/mol. Compute the boiling point of a solution containing 24.0 g of the solute and 600 g of water, when the barometric pressure is such that pure water boils at ${99.725}^{\circ }C.\left(K_{6}forwater{0.513}^{\circ }C/m\right)$

• A. ${50}^{\circ }C$
• B. ${100.08}^{\circ }C$
• C. ${150.1}^{\circ }C$
• D. ${180.34}^{\circ }C$

Answer 1

The correct option is B ${100.08}^{\circ }C$

Elevation of boiling-point describes that the boiling point of a liquid (a solvent) will be higher when another compound is added to it, meaning that a solution has a higher boiling point than a pure solvent. This happens whenever a non-volatile solute, such as a salt, is added to a pure solvent, such as water.

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

$\Delta T_b=m.K_b$

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

No. of moles of solute $=n=\frac{24}{58}=0.413$ mol

Mass of solvent in kg $=M=0.6$ kg

Molality of solute will be $=m=\frac{n}{M}=\frac{0.413}{0.6}=0.683$

Molal boiling point elevation constant is $K_b=0.513$

Elevation of boiling point can be calculated as

$\Delta T_b=m.K_b=0.683\times 0.513={0.3503}^{o}C$

${T_b}^0={99.725}^{o}C$

$T_b=99.725+0.350={100.078}^{o}C$