1. When 0.014 mol of a weak acid HA is dissolved in enough water to make 1.0 L o

Question

1. When 0.014 mol of a weak acid HA is dissolved in enough water to make 1.0 L of solution, [H+] is 1.8x10-3 M. Calculate the freezing point of this solution. Assume that the molarity concentration is equal to the molality concentration, and assume ideal behavior.  Kf (H2O) is 1.86°C kg/mol.

A) -0.058°C B) -0.029°C C) -0.026°C D) +0.029°C E) +0.026°C

ANS: B

2. You use 2.0 g of solid MX to prepare a saturated solution in 250. g of water. You find the freezing point to be –0.028°C. Calculate Ksp for the solid.
A) 2.2x 10–4

B) 5.6x 10–5

C) 1.4x10–5

D) 3.5x10–6

E) none of these

ANS: D

Explanation / Answer

Question 1-

Assuming ideal behaviour,

Depression in freezing point = Kf.m

= 1.86 X 0.014

= 0.026 Celsius

Since freezing point of water is 0 degree Celsius, freezing point of the solution would be -0.026 degree Celsius

Question 2-

Depression in freezing point = Kf.m

0.028 = 1.86 X m

m = 0.028/1.86 = 0.015

Molality of the solution = 0.015 Moles per kg

Molarity of the solution will also be 0.015 moles/liter

Ksp = [M]. [X]

= 0.015x0.015

= 2.2x 10^-4

=

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