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The boiling point of benzene is 353.23 K...

The boiling point of benzene is 353.23 K. When 1.80 g of a non-volatile solute was dissolved in 90 g of benzene, the boiling point is raised to 354.11 K. Calculate the molar mass of the solute. `K_(b)` for benzene is `2.53" K kg mol"^(-1)`.

Text Solution

Verified by Experts

The elevation of `(DeltaT_(b))` in the boiling point `=354.11" K "-353.23" K"=0.88" K"`
Substituting these values in expression we get, `M_(2)=(1000xx w_(2)xx K_(b))/(DeltaT_(b)xx w_(1))`
`M_(2)=(2.53" K kg mol"^(-1)xx1.8" g"xx1000" g kg"^(-1))/(0.88" K"xx90"g")=58" g mol"^(-1)`
Therefore, molar mass of the solute, `M_(2)=58" g mol"^(-1)`.
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