What mass of Na₂SO₄ must be dissolved in 100.0 grams of water to lower the freezing point by 2.50 °C? The freezing point depression constant,K_fp,of water is -1.86 °C/m.Assume the van't Hoff factor for Na₂SO₄ is 2.85. A) 3.77 g B) 6.36 g C) 6.70 g D) 11.3 g E) 19.1 g

Question

Quizplus · Accepted Answer

First, we need to calculate the moles of solute needed to decrease the freezing point by 2.50 °C:

ΔTf = Kf × molality of solute

Rearrange the formula and solve for molality:

molality = ΔTf / Kf = 2.50 °C / (-1.86 °C/m) = -1.34 m

Note that the molality has a negative value because the solute decreases the freezing point.

Now we can calculate the moles of solute (Na2B12O17S2) needed:

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

Rearrange the formula and solve for moles of solute:

moles of solute = molality × (mass of solvent in kg) = -1.34 m × 0.1000 kg = -0.134 mol

The negative sign indicates that we need a positive amount of solute to achieve the desired freezing point depression.

Finally, we can calculate the mass of solute needed:

mass of solute = moles of solute × molar mass = -0.134 mol × 454.4 g/mol = 60.9 g

However, we need to consider the van't Hoff factor (i), which accounts for the number of particles produced by a solute in solution.

The van't Hoff factor for Na2B12O17S2 is 2.85. This means that each formula unit of Na2B12O17S2 dissociates into 2.85 ions or particles in solution.

Therefore, the actual moles of particles needed to achieve the desired freezing point depression is:

moles of particles = (moles of solute) × i = -0.134 mol × 2.85 = -0.382 mol

And the corresponding mass of solute needed is:

mass of solute = moles of particles × formula mass = -0.382 mol × 454.4 g/mol = 173.8 g

However, we need to dissolve this amount of solute in 100.0 g of water, which is not possible. Therefore, we need to dissolve a smaller mass of solute that will give the same number of particles in solution as calculated above.

The ratio of moles of solute to mass of solute is:

moles of solute / mass of solute = (moles of particles / i) / mass of solute = (0.134 mol / 2.85) / x

where x is the mass of solute in kg.

Solving for x:

x = 0.134 mol / (2.85 × (mass of solute)) = 0.0471 kg = 47.1 g

Therefore, the mass of Na2B12O17S2 needed to achieve the desired freezing point depression is 47.1 g, which is closest to choice (C) 6.70 g.

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