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Question

Freezing point of an aqueous solution is −0.186oC. If the values of Kb and Kf of water are respectively 0.52 K kg mol−1 and 1.86 K kg mol−1 then the elevation of boiling point of the solution in K is

A
0.52
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B
1.04
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C
1.34
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D
0.134
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E
0.052
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Solution

The correct option is E 0.052∵ Freezing point of the aqueous solution =−0.186oC∴ Depression in freezing point,ΔTf=0oC−(−0.186oC) =+0.186oC=+0.186KWe know that,ΔTf=Kf⋅m ....(i)and elevation in boiling point,ΔTb=Kb⋅m ....(ii)Equation (i) Equation (ii) givesΔTfΔTb=KfKb⇒0.186KΔTb=1.86Kkgmol−10.52Kkgmol−1ΔTb=0.186×0.521.86=0.052 K

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