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)`.

Calcualate the amount of `NaCl` which must be added to `100 g` water s

1.	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» `NaCl` is a strong electrolyte. It is completely dissociated in solution. Degree of dissociation, `alpha=1` `NaClhArrunderset((n=2))(Na^o++Cl^(ө))` Number of particles after dissociation =`1+(n-1)alpha` `=1+(2-1)xx1` `=2` `DetlaT_(obs)/DetlaT_(theo)="Number of particles after dissociation"/"Number of particles when there is no dissociation"` `2/DetlaT_(theo)=2` ot `DetlaT_(theo)=1` Let W g of NaCl be dissolved in `100 g` of water. So `DetlaT_(theo)=(1000xxK_(f)xxW_(2))/(W_(1)xxMw_(1))` or`W_(2)=(DetlaT_(theo)xxW_(1)xxMw_(1))/(1000xxK_(f))` `(1xx100xx58.5)/(1000xx1.86) = 3.145 g`

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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)`.

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