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
`NaClhArrNa^++Cl^(-1)`
( n = 2 )
No. of particles after dissociation = 1 + (n – 1) `alpha`
`=1+(2-1)xx1=2`
`(DeltaT_(obs.))/(DeltaT_("theo".))=("No. of particles after dssociation")/("No. of particles when there is no dissociation")`
`2/(DeltaT_("theo."))= 2" or "DeltaT_("theo")=1`
Let w g of NaCl be dissolved in 100 g of water
`"So,"DeltaT_("theo.")=(1000xxK_fxxw)/(Wxxm)`
`"or "w=(DeltaT_("theo.")xxWxxm)/(1000xxK_f)`
`(1xx100xx58.5)/(1000xx1.86)=3.145g`