#### A solution of $\mathrm{2.5\ g}$ of a non-volatile solute in $\mathrm{50\ g}$ of water was found to boil at $\mathrm{101.4\ ^{\circ}C}$. What is the elevation in boiling point of the solution? (Molar mass of water = $\mathrm{18\ g/mol}$, Kb of water = $\mathrm{0.52\ ^{\circ}C/m}$ )Option: 1 $\mathrm{0.46\ ^{\circ}C}$Option: 2 $\mathrm{0.63\ ^{\circ}C}$Option: 3 $\mathrm{0.98\ ^{\circ}C}$Option: 4 $\mathrm{1.27\ ^{\circ}C}$

The elevation in boiling point $\Delta T_b$ can be calculated using the formula:

$\Delta T_b = i \times K_b \times m$

where i is the van't Hoff factor, $K_b$ is the molal boiling point elevation constant of the solvent, and m is the molality of the solution.

Since the solute is non-volatile, its van't Hoff factor is 1. The molality of the solution can be calculated as follows:

$Moles\ of \ solute = \frac{2.5\ g}{M_{solute}}$

$Moles\ of\ solvent = \frac{50\ g}{18\ g/mol} = 2.78\ mol$

$Molality\ (m) = \frac{\text{moles of solute}}{\text{mass of solvent (in kg)}} = \frac{\frac{2.5\ g}{M_{solute}}}{2.78\ kg} = \frac{0.142}{M_{solute}}$

Substituting the values in the formula, we get:

$\Delta T_b = 1 \times 0.52\ \mathrm{^{\circ}C/m} \times \frac{0.142}{M_{solute}} = \frac{0.0738}{M_{solute}}\ \mathrm{^{\circ}C}$

We can now solve for $M_{solute}$ using the boiling point elevation observed:

${M_{solute}}\ = \frac{0.0738}{\mathrm{1.4\ ^{\circ}C}} = 0.0527\ \mathrm{kg/mol}$

Finally, we can calculate the elevation in boiling point:

$\mathrm{101.4\ ^{\circ}C} - \mathrm{100\ ^{\circ}C} = \frac{0.0738}{M_{solute}}\ \mathrm{^{\circ}C}$

Therefore, the correct option isc) $\mathrm{0.98\ ^{\circ}C}.$

Note: The molar mass of the solute is not required to solve this problem.