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JEE · NEET Previous Year Question

$What is the boiling point of solution of 0.1 mKCl ? K_{b} of water is 0.52 K kg mol^{- 1} . (α = 100 %) (water boil at 373 K)$

Topic: JEE PYQ

Answer: B. Full step-by-step solution below.

A $100.104 K$
B $373.104 K$
C $273.104 K$
D $373.052 K$

Correct answer: B

Solution

(b) Δ T_{b} = K_{b} \times m \times i = (0.52) (0.1) (2) = 0.104 K \Rightarrow T_{b} = T_{b}^{\circ} + Δ T_{b} = 373 + 0.104 = 373.104 K

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