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Q.

‘x’ gm of solute (M. mass = 342) should be dissolved in 500gm of water so as to get a solution having difference of 105°C between freezing point and boiling point, then ‘x’ is
(Kf = 1.86Km–1, Kb = 0.52Km–1 for H2O)

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a

359

b

259

c

459

d

159

answer is C.

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Detailed Solution

∆Tb=Kbm ; ∆Tf=Kfm ∆Tb+∆Tf=m(Kb+Kf) Tb-Tbo+Tfo-Tf=WsM×1000Wo(Kb+Kf) Tb-Tf+Tfo-Tbo=x342×10005000.52+1.86 105-100=2x3422.38 x=342×52×2.38=359

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