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The boiling point of a solution of 0.11g...

The boiling point of a solution of `0.11g` of a substance is 15g of ether was found to be `0.1^(@)C` higher than that of pure ether. The molecular weight of the substance will be `(K_(b) = 2.16)`

A

148

B

158

C

168

D

178

Text Solution

Verified by Experts

The correct Answer is:
B
`m=(K_(b)xxwxx1000)/(DeltaT_(b)xxW)`
`K_(b)=2.16,w=0.11,W=15g,DeltaT_(b)=0.1`
`m=(2.16xx0.11xx1000)/(0.1xx15)=158.40~=158`.
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