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The boiling point of water becomes 100.5...

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)`).

Text Solution

Verified by Experts

The correct Answer is:
17.31 g `mol^(-1)`
`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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