1. Use elevation of boiling point relation

For a non-volatile solute, $\Delta T_b = K_b m$ where $m$ is the molality.

Since the boiling point of the two solutions is the same, their elevations in boiling point are equal: $\Delta T_{b,A} = \Delta T_{b,B}$ So, $m_A = m_B$

2. Interpret percentage solutions

A $2\%$ aqueous solution means:

• solute $A = 2\,\text{g}$
• water $= 98\,\text{g}$

An $8\%$ aqueous solution means:

• solute $B = 8\,\text{g}$
• water $= 92\,\text{g}$

3. Write molality of each solution

For solute $A$: $m_A = \frac{\text{moles of }A}{\text{kg of solvent}} = \frac{2/M_A}{98/1000}$

For solute $B$: $m_B = \frac{8/M_B}{92/1000}$

Since $m_A=m_B$, $\frac{2/M_A}{98/1000} = \frac{8/M_B}{92/1000}$

4. Simplify

$\frac{2\times 1000}{98M_A} = \frac{8\times 1000}{92M_B}$

Cancel $1000$: $\frac{2}{98M_A} = \frac{8}{92M_B}$

Cross-multiplying: $2\times 92\, M_B = 8\times 98\, M_A$ $184 M_B = 784 M_A$ $M_B = \frac{784}{184} M_A = \frac{98}{23} M_A \approx 4.26 M_A$

5. Compare with given options

This gives approximately $M_B \approx 4 M_A$ So the closest and intended option is:

Option B: $M_B = 4M_A$

6. Why exact 4 appears in standard approach

In such questions, for dilute aqueous solutions we usually take mass of solvent approximately same, so: $\Delta T_b \propto \frac{w}{M}$ Thus, $\frac{2}{M_A} = \frac{8}{M_B}$ $M_B = 4M_A$

Hence the intended answer is B.