Compute the temperature at which an aqueous solution of 4.0 g CaCl2 in 100.0 g o

Question

Compute the temperature at which an aqueous solution of 4.0 g CaCl2 in 100.0 g of water would freeze. In order to do so, determine the van't Hoff factor for CaCl2 in water.
Other info: CaCl2 solute molality/m= 0.10 solute particles per formula unit,/i = 3, & colligative molality (mc)= 0.30 Compute the temperature at which an aqueous solution of 4.0 g CaCl2 in 100.0 g of water would freeze. In order to do so, determine the van't Hoff factor for CaCl2 in water.
Other info: CaCl2 solute molality/m= 0.10 solute particles per formula unit,/i = 3, & colligative molality (mc)= 0.30
Other info: CaCl2 solute molality/m= 0.10 solute particles per formula unit,/i = 3, & colligative molality (mc)= 0.30

Explanation / Answer

we know that

moles = mass / molar mass

molar mass of CaCl2 = 111 g /mol

so

moles of Cacl2 = 4 / 111

now

molality = moles of CaCl2 x 1000 / mass of water (g)

so

molality = (4/111) x 1000 / 100

molality = 0.36036

now

we know that

depression in freezing point is given by

dTf = i x Kf x m

now

CaCl2 --> Ca+2 + 2 Cl-

so 3 particles after dissociation

so i = 3

Kf for water = 1.86

so

using those values

dTf = 3 x 1.86 x 0.36036

dTf = 2.0108

now

we know that

dTf = freezing point of solvent water - freezing point of solution

we know that

freezing point of water= 0

so

2.0108 = 0 - freezing point of solution

freezing point of solution= -2.0108

so

the freezing point of the given solution is -2.0108 C

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