Question

Calculate the freezing point of a one molar aqueous solution (density = 104 g L') of KCI. (Kb for
water = 1.86 kg mol', atomic masses of K = 39, C1 = 35.5).
K for​

Answer 1

Explanation:

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Calculate the freezing point of a molar aqueous solution of KCl. (Density of solution =1.04gml

−1

K

f

=1.86 K kgmol

−1

)

Medium

Solution

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Molarity of KCl solution =1 M

This means 1 mole (or 74.5 g) of KCl is dissolved in 1 L (or 1000 ml) of solution.

Mass of solution=density of solution×volume of solution

Mass of solution=1.04×1000=1040 g

Mass of solvent=mass of solution−mass of solute=1040−74.5=965.5 g=0.9655 kg

Molality of solution(m)=

mass of solvent in kg

moles of solute

m=

0.9655

1

=1.035 mol kg

−1

ΔT

f

=K

f

.m

(0

0

C)−(T

f

)=1.86×1.035

T

f

=−1.92

0

C=271.08 K