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))`

Options

A. `-0.570^(@)C`
B. `-0.372^(@)C`
C. `-0.520^(@)C`
D. `+0.372^(@)C`

Answer

2

Explanation

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`