1.

The boliling point of 5% solution (W/W) of non-volatile solute in water is `100.45^(@)C`. The boiling point of pure water is `100^(@)C`.Calculate th molar mas of th solute `(K_(b)"for water=0.52 K kg mol"^(-1))`.

Answer» Correct Answer - 60.8 g `mol^(-1)`
`DeltaT_(b)=100.45^(@)C-100^(@)C=0.45 K`
`W_(B)=5g, W_(A)=100-5=95g, 0.095 kg, K_(b)=0.52" K kg mol"^(-1)`
`M_(B)=(K_(b)xxW_(B))/(DeltaT_(b)xxW_(A))=((0.52K" kg mol"^(-1))xx(5g))/((0.45K)xx(0.095 kg))=60.8" g mol"^(-1)`.


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