Q

# 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

535 Views

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.

Exams
Articles
Questions