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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 `(Delta T_(b))` in the boiling point `= 354.11 K - 353.23 K = 0.88 K`
Sdubstituting these values in expression
`M_(2)=(1000 xx w_(2))xx (K_(b))/(Delta T_(b)xx w_(1))`
we get,
`M_(2)=(2.53"K kg mol"^(-1)xx1.8 g xx1000 g kg^(-1))/(0.88K xx 90 g)`
`= 58 g mol^(-1)`
Therefore, molar mass of the solute,
`M_(2)=58 g mol^(-1)`
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