Boiling point of a solution containing ‘x’ grams of KCl ( $α$ = 1) in 500 grams of H₂O is 101.04°C. Value of ‘x’ is (K_b of H₂O = 0.52K kg mole^-1 and molar mass of KCl = 74.5 g/mole)

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55.875 g

74.5 g

37.25 g

18.625 g

answer is C.

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Detailed Solution

$\begin{array}{l} {ΔT}_{b} = 0 . 52^{\circ} \\ {ΔT}_{b} = {iK}_{b} m \\ 1.04 = 2 (0.52) \times \frac{x}{74.5} \times \frac{1000}{500} \\ x = \frac{74.5}{2} = 37.25 g \end{array}$

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