1. What will be the freezing point of water an aqueous solution containing 80g of NaCl?
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
"\u2206T= 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
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