1. When 0.014 mol of a weak acid HA is dissolved in enough water to make 1.0 L o
ID: 573281 • Letter: 1
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
3.Which of the following statements is(are) true?
A) The rate of dissolution of a solid in a liquid always increases with increasing temperature.
B) The solubility of a solid in a liquid always increases with increasing temperature.
C) According to Henry's law, the amount of gas dissolved in a solution is directly proportional to the pressure of the gas above the liquid.
D) Two of these statements are true.
E) All of these statements are true.
P.sI Ask Q1 and Q2 on Chegg before, but the answers are wrong. Thank you for your help.
Explanation / Answer
Ans 1 : B) -0.029oC
The depression in freezing point is given as : kf . m
Number of moles of [H+] + [A-] = 1.8 x 10-3M x 2 = 3.6 x 10-3
So number of moles of undissociated acid = 0.014 - (1.8 x 10-3) = 0.0122 mol
total no. of moles = 0.0122 + 3.6 x 10-3 = 0.0158 mol
Molarity = no. of moles of acid / volume of solution in L
= 0.0158 / 1
= 0.0158 M
since here molarity and molality concentrations are equivalent , we will substitute it for the formula given above :
Depression in freezing point = 1.86 x 0.0158
= 0.029o C
Since the freezing point of pure water is 00C , so the freezing point of the solution = 0oC -0.029oC
= - 0.029oC
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