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The boiling point of benzene is 353.23 K. When 1.80 g of a non-volatile solute was dissolved in 90 g of benzene, the boiling point is raised to 354.11 K. Calculate the molar mass of the solute. K_(b) for benzene is 2.53 K kg mol^(-1). |
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Answer» Solution :Here, we are GIVEN `"" w_(2)=1.80 g, w_(1)=90 g, "" Delta T_(B)=354.11-353.23 K=0.88 K` `"" K_(b)=2.53 " K kg mol"^(-1)` Substituting these VALUES in the formula, `M_(2)=(1000 K_(b)w_(2))/(w_(1)Delta T_(b))`, we get `M_(2)=(1000 " g kg"^(-1)xx2.53" K kg mol"^(-1)xx1.80 g)/(90g XX 0.88 K)=58 "g mol"^(-1)` |
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