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In a 0.2 molal aqueous solution of a weak acid HX the degree of ionization is 0.3 .Taking $K_{f}$ for water as 1.85 the freezing point of the solution will be nearest to

Option 1)
$0.480^{\circ}C$ $0.480^{\circ}C$ $0.480^{\circ}C$ $-0.480^{\circ}C$

Option 2)
$-0.360^{\circ}C$

Option 3)
$-0.260^{\circ}C$

Option 4)
$+0.480^{\circ}C$

Answers (1)

best_answer

As we learnt in

Application of Vant Hoff factor -

$(a)\: \: \: \left ( \frac{\Delta P}{P^{0}} \right )_{obs}= I_{solute}\times i$

$(b)\: \: \: \left ( \Delta T_{b} \right )_{obs}= K_{b}\times m\times i$

$(c)\: \: \: \left ( \Delta T_{f} \right )_{obs}= K_{f}\times m\times i$

$(d)\: \: \: \left ( \pi \right )_{obs}= C\times R\times T\times i$

-

$HX\rightleftharpoons H^{+}+X^{-}$

$1\: \: \: \: \: \: \: \: \: \: \: \: \: 0\: \: \: \: \: \: 0$

$1-0.3\: \: \: \: \: \: \: \: 0.3\: \: \: 0.3$

Total number of moles after dissociation $= 1-0.3+0.3+0.3= 1.3$ = i

$\Delta T_{f}=i K_{f}\times m$

$\Delta T_{f}=1.3\times 1.85\times 0.2=0.4810$

Freezing point of solution = 0 - 0.4810 = - 0.4810 c^o.

Correct option is 1.

Option 1)

$– 0.480^{\circ}C$ $– 0.480^{\circ}C$ $– 0.480^{\circ}C$ $-0.480^{\circ}C$

This is the correct option.

Option 2)

$-0.360^{\circ}C$

This is an incorrect option.

Option 3)

$-0.260^{\circ}C$

This is an incorrect option.

Option 4)

$+0.480^{\circ}C$

This is an incorrect option.

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