A solution containing 0.90 g of a non-volatile solute in 100 mL of benzene having density 0.88 g `mL^(-1)` at 298 K, boils at a temperature `0.25^(@)`

Question

A solution containing 0.90 g of a non-volatile solute in 100 mL of benzene having density 0.88 g `mL^(-1)` at 298 K, boils at a temperature `0.25^(@)`

A solution containing 0.90 g of a non-volatile solute in 100 mL of benzene having density 0.88 g `mL^(-1)` at 298 K, boils at a temperature `0.25^(@)` higher than benzne. Calculate the molecular mass of solute. Molal boiling point elevation constant for benzenene is 2.52 K kg `mol^(-1)`.

Monjur Alahi

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2025-01-04 04:42

1 Answers

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Salim Ahmed Kawsar · Accepted Answer · 2023-02-05T15:29:48+06:00

Salim Ahmed Kawsar

Correct Answer - 103.1 g `mol^(-1)`
`T_(b)^(@)=78^(@)C,T_(b)=78.28^(@)C,DeltaT_(b)=T_(b)-T_(b)^(@)=78.28-780.28^(@)C=0.28 K`
`K_(b)=1.15" K kg mol"^(-1), W_(B)=1.12 g, W_(A)=0.032 kg, M_(B)= ?`
`M_(B)=(K_(b)xxW_(B))/(DeltaT_(b)xxW_(A))=((1.5" K kg mol"^(-1))xx(1.12 g))/((0.28 K)xx(0.032 Kg ))=143.75" g mol"^(-1)`.

9 views Answered 2023-02-05 15:29