The freezing point depression constant for water is `-1.86^(@)Cm^(-1).`if `5.00g Na_(2)SO_(4)`is dissolved in `45.0gH_(2)O`,the freezing point is changed by `-3.82^(@)C`,Calculate the van't Hoff factor for `Na_(2)SO_(4)`

Options

A. 0.381
B. 2.05
C. 2.63
D. 3.11

Answer

3

Explanation

We have
`DeltaT_(f)=iK_(f)m`
where m is molality of solution. Thus
`i=(DeltaT_(f))/(K_(f)m)=(DeltaT_(f))/(K_(f)w_(solute)/(M_(solute)w_(solvent)))`
We are given
`DeltaT_(f)=T_(f)^(0)-T_(f)=(0.00^(@)C)-(-3.82^(@)C)`
`=3.82^(@)C`
`K_(f)=1.86^(@)C m^(-1)`
`W_(solute)=5.00 g`
`M_(solute)=(23xx2+32xx1+16xx4)`
`=142 g mol^(-1)`
`W_(solvent)=45.0 g=0.045 kg`
Substituting these results, we get
`i=(3.82^(@)C)/(1.86^(@)C m^(-1)(5.00 g)/(142 g mol^(-1)xx0.045 g))`
`=2.624`
`=2.63`