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]
Answer (integer)
100
Solution
$\Delta$T<sub>b</sub> = Boiling point of the solution $-$ Boiling point of the pure solvent<br><br>= 100.52 $-$ 100<br><br>= 0.52<br><br>$\therefore$ $\Delta$T<sub>b</sub> = K<sub>b</sub> (iM)<br><br>$\Rightarrow$ 0.52 = i $\times$ 0.52 $\times$ 2<br><br>$\Rightarrow$ i = ${1 \over 2}$<br><br>We know, <br><br>i = 1 + $\left( {{1 \over n} - 1} \right)\beta$<br><br>here, $\beta$ = degree of dimerization<br><br>n = number of particle associated<br><br>n = 2 for dimerization<br><br>$\therefore$ ${1 \over 2} = 1 + \left( {{1 \over 2} - 1} \right)\beta$<br><br>$\Rightarrow \beta = 1$<br><br>$\therefore$ % association = 100
About this question
Subject: Chemistry · Chapter: States of Matter · Topic: Gas Laws
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