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A solution of sucrose (molar mass `342g mol^(-1)`has been produced by dissolving`68.5g`sucrose in `1000g`water .The freezing point of the solution obtained will be `(K_(f)` for `H_(2)O=1.86kgmol^(-1))`A. `-0.570^(@)C`B. `-0.372^(@)C`C. `-0.520^(@)C`D. `+0.372^(@)C` |
Answer» Correct Answer - 2 We have `DeltaT_(f)=iK_(f)m` `i=1` for sucross as it neither associates nor dissociates `K_(f)=1.86 K kg mol^(-1)` `m=n_("sucrose")/(Kg_(H_(2)O))=(68.5 g//342 g mol^(-1))/(1 kg)` `=0.2 mol kg^(-1)` Substituting these result, we get `DeltaT_(f)=(1)(1.86 K kg mol^(-1))(0.2 mol kg^(-1))` `=0.372K or 0.372^(@)C` Freezing point of solution `=("Freezing point of water")-(DeltaT_(f))` `=(0.000^(@)C)-(0.372^(@)C)` `=-0.372^(@)C` |
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