A solute A dimerizes in water- The boiling point of a 2 molal solution of A is 100-52 -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 -C-

Molality, m = 2; T_b of solution = 100.52 ^oC ΔT_b = 0.52 ΔT_b= i × K_b × m 0.52 = i × 0.52 × 2 i = 0.5 Degree of association (β) : β=(i 1)× n(1 n) Since ...

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Standard XII

Chemistry

Colligative Properties

A solute A di...

Question

A solute A dimerizes in water. The boiling point of a 2 molal solution of A is 100.52 ºC. The percentage association of A is________.(Round off to the nearest integer.)

[Use : $K_{b}$ for water = $0.52 K k g m o l^{- 1}$
Boiling point of water = 100 ºC]

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Solution

$molality, m = 2; T_{b} of solution = 100.52^{o} C$

$Δ T_{b} = 0.52$

$Δ T_{b} = i \times K_{b} \times m$

$0.52 = i \times 0.52 \times 2$
$i = 0.5$

Degree of association $(β)$ :
$β = \frac{(i - 1) \times n}{(1 - n)}$
Since solute A dimerises, $n = 2$
$β = \frac{(0.5 - 1) \times 2}{(1 - 2)}$
$β = 1$
Percentage of degree of association is 100%

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A solute A dimerizes in water. The boiling point of a 2 molal solution of A is 100.52 ºC. 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 ºC]

molality, m=2;Tb of solution =100.52 oC ΔTb=0.52 ΔTb=i×Kb×m 0.52=i×0.52×2 i=0.5 Degree of association (β) : β=(i−1)×n(1−n) Since solute A dimerises, n=2 β=(0.5−1)×2(1−2) β=1 Percentage of degree of association is 100%

A solute A dimerizes in water. The boiling point of a 2 molal solution of A is 100.52 ºC. The percentage association of A is________.(Round off to the nearest integer.)

[Use : $K_{b}$ for water = $0.52 K k g m o l^{- 1}$
Boiling point of water = 100 ºC]