SolutionHard
Question
64.4g Crystals of pure Na2SO4 10 H2O are dissolved in 114 g of water to obtain an aq. solution of salt. Determine the freezing point of solution assuming that salt is completely ionized in solution Kf of water is 1.86 K kgmol-1 :-
Options
A.-7.44oC
B.-9.79oC
C.-3.26oC
D.-2.48oC
Solution
Moles of solute (n) = 
= 0.2 mol Na2SO4 10 H2O
∵ 1 mol Na2SO4 10H2O contains = 10 mol H2O
⇒ 180 gH2O
∴ 0.2 mol Na2SO4 10H2O contains = 180 × 0.2 = 36 g H2O
Total mass of solvent (H2O) in solu. = 114 + 36 = 150 g
∴ ᐃTf = 3 × 1.86 ×
= 7.44
= 0.2 mol Na2SO4 10 H2O∵ 1 mol Na2SO4 10H2O contains = 10 mol H2O
⇒ 180 gH2O
∴ 0.2 mol Na2SO4 10H2O contains = 180 × 0.2 = 36 g H2O
Total mass of solvent (H2O) in solu. = 114 + 36 = 150 g
∴ ᐃTf = 3 × 1.86 ×
= 7.44Create a free account to view solution
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