The elevation of boiling point of a solution of a non-volatile solute is written in terms of the following relations:
\( \displaystyle \Delta T_b = T_b - T_b^0 \qquad \text{and} \qquad m = \frac{1000 \times w_2}{M_2 \times w_1} \)
Match the symbols in List-I with what they stand for in List-II.
| List-I (Symbol) | List-II (Stands for) |
|---|---|
| (A) Tb0 | (I) Mass of the solvent in grams |
| (B) w2 | (II) Boiling point of the solution |
| (C) Tb | (III) Mass of the solute in grams |
| (D) w1 | (IV) Boiling point of the pure solvent |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → A-IV, B-III, C-II, D-I
Two conventions fix every symbol. The superscript 0 marks the pure solvent, and in a binary solution the subscript 1 stands for the solvent and 2 for the solute. A solution of a non-volatile solute boils at a higher temperature than the pure solvent, so ΔTb = Tb − Tb0 is positive.
(A) Tb0 — This matches (IV) Boiling point of the pure solvent. It is the lower of the two temperatures.
(B) w2 — This matches (III) Mass of the solute in grams. It goes with M2, the molar mass of the solute, to give the moles of solute, w2/M2.
(C) Tb — This matches (II) Boiling point of the solution. It is the higher temperature.
(D) w1 — This matches (I) Mass of the solvent in grams. Dividing it by 1000 gives the kilograms of solvent, which is what molality needs.
NCERT (2026-27 edition):
1.6.2 Elevation of Boiling Point – “Let Tb0 be the boiling point of pure solvent and Tb be the boiling point of solution. The increase in the boiling point ΔTb = Tb − Tb0 is known as elevation of boiling point.”
“If w2 gram of solute of molar mass M2 is dissolved in w1 gram of solvent, then molality, m of the solution is given by the expression:”
\( \displaystyle m = \frac{w_2/M_2}{w_1/1000} = \frac{1000 \times w_2}{M_2 \times w_1} \qquad (1.31) \)