1.

The boiling point of water becomes 100.52^(@)C if 1.5 g of a non-volalite solute is dissolved in 100 mL of it. Calculate the molecular weight of the solute. (K_(b) for water=0.6 K kg mol^(-1)). (density of waer-1 g moL^(-1)).

Answer»


Solution :`DeltaT_(b)=100.52-100=0.52^(@)C=0.52 K`
`W_(B)=1.5g, W_(A)=100g=0.1 kg ""[1" mL of water"~~1G]`
`K_(b)=0.6K//m, M_(B)=?`
`M_(B)=(K_(b)xxW_(B))/(DeltaT_(b)xxW_(A))=((0.6" K kg MOL"^(-1))xx(1.5g))/((0.52 K)xx(0.1 kg))=17.31" g mol"^(-1)`


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