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Please help! A 0.1 molal aqueous solution of a weak acid is 30% ionized. If Kf for water is 1.86°C/m , the freezing point of the solution will be:

A 0.1 molal aqueous solution of a weak acid is 30% ionized. If Kf for water is 1.86°C/m , the freezing point of the solution will be:

  • Option 1)

    -0.18°C

  • Option 2)

    -0.54°C

  • Option 3)

    -0.36°C

  • Option 4)

    -0.24°C3

 
Answers (1)
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As we learned in concept

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

-

 

 (\Delta T_{F})_{obs}\:=\: Kf \times m \times i

                    = 1.86\times 0.1\times 1.3=0.2418\:C^{\circ}

 


Option 1)

-0.18°C

This option is incorrect.

Option 2)

-0.54°C

This option is incorrect.

Option 3)

-0.36°C

This option is incorrect.

Option 4)

-0.24°C3

This option is correct.

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