Question

What is the freezing point of a solution containing 8.1 g HBr
in 100 g water assuming the acid to be 90% ionised ?
(Kf for water = 1.86 K kg mol⁻¹) :
(a) 0.85K (b) – 3.53K(c) 0K (d) – 0.35K

Answer 1

answer : option (b) -3.53K

given, mass of HBr = 8.1g

molecular weight of HBr = 81 g/mol

so, no of mole of HBr = 8.1/81 = 0.1

mass of water = 100g

so, molality , m = no of mole of solute/mass of solvent in Kg

= 0.1/(100/1000 ) molal

= 0.1 × 10 = 1 molal.

dissociation reaction of HBr....

HBr ⇔H^+ + Br^-

at eql, 1 - α α α

i = (1 - α) + α + α = 1 + α

as it is given, acid is to be 90% ionised.

so, α = 0.9

now, i = 1 + 0.9 = 1.9

using formula, T_f = -i × k_f × m

= -1.9 × 1.86 × 1

= -3.534 K

hence option (b) is correct choice.

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