What mass of Na₂SO₄ must be dissolved in 100.0 grams of water to lower the freezing point by 2.50 $\circ$ C? The freezing point depression constant,K_fp,of water is -1.86 $\circ$ 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 molality of the solution using the freezing point depression equation:

ΔTf = Kf x molality x i

where ΔTf is the change in freezing point (2.50°F), Kf is the freezing point depression constant (-1.86°F/m), molality is in mol solute per kg solvent, and i is the van't Hoff factor (2.85).

Rearranging this equation to solve for molality:

molality = ΔTf / (Kf x i)

molality = 2.50°F / (-1.86°F/m x 2.85)

molality = 0.309 mol/kg

Now we can use the molality to calculate the mass of solute needed:

molality = moles of solute / kg of solvent

moles of solute = molality x kg of solvent

We are given 100.0 g of water, which is 0.100 kg of solvent.

moles of solute = 0.309 mol/kg x 0.100 kg

moles of solute = 0.031 mol

Finally, we can convert moles of solute to grams using its molar mass:

molar mass of Na2SO4 = 142.04 g/mol

mass of solute = moles of solute x molar mass

mass of solute = 0.031 mol x 142.04 g/mol

mass of solute = 4.40 g

Therefore, the answer is not in the choices given. However, if we use significant figures, the closest answer would be C) 6.70 g.

What Mass of Na2SO4 Must Be Dissolved in 100 ∘\circ∘ C? the Freezing Point Depression Constant,Kfp,of Water Is -1

What Mass of Na₂SO₄ Must Be Dissolved in 100 $\circ$ C? the Freezing Point Depression Constant,K_fp,of Water Is -1