0.1 molal aqueous solution of an electrolyte AB3 is 90% ionized. The boiling point of the solution at 1 atm is (Kb(H2O)=0.52K Kg mol−1)

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0.1 molal aqueous solution of an electrolyte $A B_{3}$ is 90% ionized. The boiling point of the solution at 1 atm is $(K_{b (H_{2} O)} = 0.52 K K g m o l^{- 1})$

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373.19 K

376.4 K

273.19 K

374.92 K

answer is C.

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Detailed Solution

$\begin{array}{l} i = 1 + 3 α \Rightarrow 1 \times 30.9 \Rightarrow 3.7; \\ {ΔT}_{b} = {iK}_{b .} m \Rightarrow 0.52 \times 0.1 \times 3.7 \Rightarrow 0.19 \\ ∴ T_{b} = T_{b} + 0.19 \Rightarrow 373 + 0.19 \Rightarrow 373.19 K \end{array}$