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The boiling point of water is 100^(@)C a...

The boiling point of water is `100^(@)C` and it becomes `100.52^(@)C` if 3 g of a non-voltile is dissolved in 20 mL of it. Calculate the molecular weight of the solute. (`K_(b)` for water is 0.52 K kg `mol^(-1)`, density of water= 1 g `mol^(-1)`).

Text Solution

Verified by Experts

The correct Answer is:
150 g `mol^(-1)`
`DeltaT_(b)=100.52-100=0.52^(@)C=0.52 K`
`W_(B)= 3 g, W_(A)=20 g [20" mL of water= 20 g"], 0.02 kg, k_(b)=0.52" K kg mol"^(-1)`
`M_(B)= ?`
`M_(B)=(K_(b)xxW_(b))/(DeltaT_(b)xxW_(A))=((0.52" K kg mol"^(-1))xx(3g))/((0.52K)xx(0.02 kg))=150" g mol"^(-1)`.
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