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)`

Options

A. 148
B. 158
C. 168
D. 178

Answer

B

Explanation

`M_(2) = (1000 xx K xx W_(2))/(Delta T_(B) xx W_(1))`
` = (1000 xx 2.16 xx 0.11)/(0.1 xx 15) = 158`