JEE Chemistry Practice Question

The boiling point of water in a 0.1 molal silver nitrate solution (solution $\mathbf{A}$ ) is $\mathbf{x}^{\circ} \mathrm{C}$. To this solution $\mathbf{A}$, an equal volume of 0.1 molal aqueous barium chloride solution is added to make a new solution $\mathbf{B}$. The difference in the boiling points of water in the two solutions $\mathbf{A}$ and $\mathbf{B}$ is $\mathbf{y} \times 10^{-2}{ }^{\circ} \mathrm{C}$. (Assume: Densities of the solutions $\mathbf{A}$ and $\mathbf{B}$ are the same as that of water and the soluble salts dissociate completely. Use: Molal elevation constant (Ebullioscopic Constant), $K_b=0.5 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1}$; Boiling point of pure water as $100^{\circ} \mathrm{C}$.) What is the value of $\mathbf{x}$?