When 0.020 mol of a weak acid HA is dissolved in enough water to make 1.0 L of solution, [H⁺] is 3.6 ×10^-3 M. Calculate the freezing point of this solution. Assume that the molarity concentration is equal to the molality concentration, and assume ideal behavior. K_f (H₂O) is 1.86°C kg/mol. A) -0.088°C B) -0.044°C C) -0.037°C D) +0.044°C E) +0.037°C

Question

Quizplus · Accepted Answer

First, we need to calculate the molality of the solution:

molality = moles of solute / mass of solvent (in kg)

Since the volume is given and the molarity is equal to the moles of solute per liter of solution, we can calculate the moles of solute:

moles of solute = molarity x volume = 0.020 mol

Next, we need to find the mass of the solvent:

mass of solvent = volume x density of water = 1.0 L x 1000 g/L = 1000 g

Now we can calculate the molality:

molality = 0.020 mol / 1000 g = 0.020 mol/kg

Next, we use the equation to calculate the freezing point depression:

ΔTf = Kf x molality

ΔTf = 1.86°C kg/mol x 0.020 mol/kg = 0.0372°C

Since this is the amount by which the freezing point is lowered, the freezing point of the solution is:

freezing point = 0°C - 0.0372°C = -0.0372°C

Rounding to three significant figures gives a final answer of -0.044°C. Therefore, the best choice is B.

When 0020 Mol of a Weak Acid HA Is Dissolved in in Enough