JEE MAIN - Chemistry (2021 - 18th March Evening Shift - No. 22)
A solute A dimerizes in water. The boiling point of a 2 molal solution of A is 100.52$$^\circ$$C. The percentage association of A is __________. (Round off to the Nearest Integer).
[Use : Kb for water = 0.52 K kg mol$$-$$1 Boiling point of water = 100$$^\circ$$C]
[Use : Kb for water = 0.52 K kg mol$$-$$1 Boiling point of water = 100$$^\circ$$C]
Answer
100
Explanation
$$\Delta$$Tb = Boiling point of the solution $$-$$ Boiling point of the pure solvent
= 100.52 $$-$$ 100
= 0.52
$$ \therefore $$ $$\Delta$$Tb = Kb (iM)
$$ \Rightarrow $$ 0.52 = i $$\times$$ 0.52 $$\times$$ 2
$$ \Rightarrow $$ i = $${1 \over 2}$$
We know,
i = 1 + $$\left( {{1 \over n} - 1} \right)\beta $$
here, $$\beta$$ = degree of dimerization
n = number of particle associated
n = 2 for dimerization
$$ \therefore $$ $${1 \over 2} = 1 + \left( {{1 \over 2} - 1} \right)\beta $$
$$ \Rightarrow \beta = 1$$
$$ \therefore $$ % association = 100
= 100.52 $$-$$ 100
= 0.52
$$ \therefore $$ $$\Delta$$Tb = Kb (iM)
$$ \Rightarrow $$ 0.52 = i $$\times$$ 0.52 $$\times$$ 2
$$ \Rightarrow $$ i = $${1 \over 2}$$
We know,
i = 1 + $$\left( {{1 \over n} - 1} \right)\beta $$
here, $$\beta$$ = degree of dimerization
n = number of particle associated
n = 2 for dimerization
$$ \therefore $$ $${1 \over 2} = 1 + \left( {{1 \over 2} - 1} \right)\beta $$
$$ \Rightarrow \beta = 1$$
$$ \therefore $$ % association = 100
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