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How many grams of sucrose (molecular weight 342) should be dissolved in `100 g` water in order to produce a solution with `105^(@)C` difference between the freezing point and the boiling point ? `(K_(b) =0.51^(@)C m^(-1) , (K_(f) =1.86^(@)C m^(-1))`A. `34.2g`B. `72g`C. `342g`D. `460g` |
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Answer» Correct Answer - B `"Boiling point "(T_(b))=100+DeltaT_(b)=100+k_(b)m` `"Freezing point "(T_(f))=0-DeltaT_(f)=-k_(f)m` `T_(b)-T_(f)=(100+k_(b)m)-(-k_(f)m)` `105=100+0.51m+1.86m` `2.37m=5" or m"=(5)/(2.37)=2.11` `:."Weight of sucrose to be dissolved in 100g water"` `=(2.11xx342)/(1000)xx100=72g` |
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