Solutions Question 53

Question: The boiling point of water ( 100oC ) becomes 100.52oC , if 3 grams of a nonvolatile solute is dissolved in 200ml of water. The molecular weight of solute is ( Kb for water is 0.6,K−m ) [AIIMS 1998]

Options:

A) 12.2,g,mol−1

B) 15.4,g,mol

C) 17.3,g,mol−1

D) 20.4,g,mol

Show Answer

Answer:

Correct Answer: C

Solution:

First boiling point of water = 100oC Final boiling point of water = 100.52o

w=3g , W=200,g , Kb=0.6,kg−1

ΔTb=100.52−100=0.52oC

m=Kb×w×1000ΔTb×W

=0.6×3×10000.52×200=1800104=17.3gmol−1 .

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