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)`.

Explanation

NaCl is a strong electrolyte. It is completely dissociated in solution
Degree of dissociation, `alpha =1`
`NaClhArr underset((n=2))(Na^(+)+Cl^(-))`
No. of particles after dissociation `= 1+(n-1)alpha`
`=1+(2-1)xx1=2`
`(DeltaT_(obs))/(DeltaT_(theo))=("No. of particles after dissociation")/("No. of particles when there is no dissociation")`
`(2)/(DeltaT_(theo))=2`
or `DeltaT_(theo)=1`
Let 2g of NaCl be dissolved in 100g of water
So, `DeltaT_(theo)=(1000xxK_(f)xxw)/(Wxxm)`
or `w=(DeltaT_(theo)xxWxxm)/(1000xxK_(f))=(1xx100xx58.5)/(1000xx1.86)`
`=3.145g`.