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)`).

The boiling point of water becomes `100.52^(@)C` if 1.5 g of a non-vol

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» Correct Answer - 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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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)`).

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