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(B) For a dilute aqueous solution,the elevation in boiling point is given by $\Delta T_B = K_B 	imes m$,where $m$ is the molality.Since the boiling p

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$M_B = 4M_A$

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(B) For a dilute aqueous solution,the elevation in boiling point is given by $\Delta T_B = K_B 	imes m$,where $m$ is the molality.Since the boiling points are equal,the elevation in boiling point is the same,so the molalities must be equal: $m_A = m_B$.For $A$: $2 \ g$ of solute in $98 \ g$ of solvent. Molality $m_A = \frac{2 / M_A}{98 / 1000} = \frac{2000}{98 M_A}$.For $B$: $8 \ g$ of solute in $92 \ g$ of solvent. Molality $m_B = \frac{8 / M_B}{92 / 1000} = \frac{8000}{92 M_B}$.Equating $m_A = m_B$: $\frac{2000}{98 M_A} = \frac{8000}{92 M_B}$.$\frac{1}{98 M_A} = \frac{4}{92 M_B}$.$M_B = \frac{4 	imes 98}{92} M_A = \frac{392}{92} M_A \approx 4.26 M_A$.Given the standard approximation often used in such problems where the mass of solvent is taken as $100 \ g$ (assuming dilute solution),the ratio becomes $\frac{2/M_A}{100} = \frac{8/M_B}{100}$,which leads to $M_B = 4M_A$.

Boiling point of a $2\%$ aqueous solution of a non-volatile solute $A$ is equal to the boiling point of an $8\%$ aqueous solution of a non-volatile solute $B$. The relation between molecular weights of $A$ and $B$ is:

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