Calcualate the amount of `NaCl` which must be added to `100 g` water so that the freezing point, depressed by `2 K`. For water `K_(f)` =`1.86 K kg mol^(-1)`.

Answer

3.147 g

Explanation

`NaCl rarrNa^(+)+Cl^(-),i=2, DeltaT_(f)=eK_(f)m, i.e., 2=2xx1.86xxm or m=0.538`
`therefore" Amount of NaCl to bedissolved in 100 g"=(0.538xx100)/(1000)xx58.5g=3.147g.`