Question

1.      What will be the freezing point of water an aqueous solution containing 80g of NaCl?

Answer 1

Kf=1.86 C⁰/m

let the mass of water be 160 g

Find the amount of substance = contained in 160 g of water by the formula

$n = \frac{m} {M}=\frac{80}{58.443}=1.369$

Find the amount of substance contained in 1000 g of water:

$n =\frac{1.3698\times1000}{160} =8.55$

We find a decrease in the freezing temperature of the solution by the formula

$∆T= Kcr\times C=1.86\times8.55=15.903$

∆Tsam. = Kcr. * C

where C is the molar concentration of the solution

Find the freezing point of the solution by the formula

FP = 0 – 15.903= - 15.903К

or -289 C⁰