MHTCET 11 Apr 2026 Shift 2 Paper

Concept:Boiling point elevation is a colligative property that depends on the number of solute particles, allowing molar mass to be determined using $\Delta T_b = K_b \times m$.Explanation:The elevation in boiling point is given by:$$\Delta T_b = T_b - T_b^\circ$$Substitute the given boiling point and pure solvent boiling point:$$\Delta T_b = 357.27 - 350.06 = 7.21 \ \text{K}$$Using the boiling point elevation formula:$$\Delta T_b = K_b \times m$$Rearrange to find molality $m$:$$m = \frac{\Delta T_b}{K_b} = \frac{7.21}{2.77} \approx 2.60 \ \text{mol/kg}$$Convert the mass of solvent Y to kg:$$150 \ \text{g} = 0.150 \ \text{kg}$$Calculate the moles of solute X:$$\text{Moles of solute} = m \times \text{mass of solvent in kg}$$$$\text{Moles of solute} = 2.60 \times 0.150 = 0.390 \ \text{mol}$$Now calculate the molar mass:$$\text{Molar mass} = \frac{\text{mass of solute}}{\text{moles of solute}}$$$$\text{Molar mass} = \frac{50}{0.390} \approx 128 \ \text{g/mol}$$Answer:The molar mass of solute X is $128 \ \text{g/mol}$, so the correct option is B.