To observe an elevation of boiling point of `0.05^(@)C`, the amount of a solute (molecular weight = 100) to be added to 100 g of water `(K_(b) = 0.5)` isA. 2 gB. 0.05 gC. 1 gD. 0.75 g

To observe an elevation of boiling point of `0.05^(@)C`, the amount of

1.	To observe an elevation of boiling point of `0.05^(@)C`, the amount of a solute (molecular weight = 100) to be added to 100 g of water `(K_(b) = 0.5)` isA. 2 gB. 0.05 gC. 1 gD. 0.75 g
Answer» Correct Answer - c Elevation of boiling point, `DeltaT_(b) = (wxxK_(b)xx1000)/(MxxW_("gram"))` On substituting values, we get `0.05 = (wxx0.5xx1000)/(100xxx100)` or `w = (0.05xx100xx100)/(0.5xx1000) = 1g`

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To observe an elevation of boiling point of `0.05^(@)C`, the amount of a solute (molecular weight = 100) to be added to 100 g of water `(K_(b) = 0.5)` isA. 2 gB. 0.05 gC. 1 gD. 0.75 g

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