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)

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 co

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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