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A solution of 12.5 g of an unknown solut...

A solution of 12.5 g of an unknown solute in 170 g water gave a boiling point elevation of 0.63 K. Calculate the molar mass of the solute. `(K_(b)=0.53 K kg mol^(-1))`.

Text Solution

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The correct Answer is:
61.85 g `mol^(-1)`
`W_(B)=12.5g, W_(A)=0.170 kg, DeltaT_(b)=0.63 K, K_(b)=0.53" K kg mol"^(-1), M_(B)=?`
`M_(B)=(K_(b)xxW_(B))/(DeltaT_(b)xxW_(A))=((0.53" K kg mol"^(-1))(12.5g))/((0.63 K)xx(0.170 kg))=61.85" g mol"^(-1)`.
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