You have a 10.40-g mixture of table sugar (C₁₂H₂₂O₁₁) and table salt (NaCl). When this mixture is dissolved in 150. g of water, the freezing point is found to be -2.24°C. Calculate the percent by mass of sugar in the original mixture. (Assume that the NaCl is completely dissociated.) A) 61.0% B) 53.8% C) 46.2% D) 39.0% E) none of these

Question

Quizplus · Accepted Answer

First, we need to calculate the molality of the solution using the freezing point depression equation:

ΔTf = Kf · molality

where ΔTf = 2.24°C, Kf = 1.86°C/m (the freezing point depression constant of water), and molality = moles of solute per kilogram of solvent.

Let x be the number of moles of table sugar in the mixture. Since the molecular weight of table sugar is 342.3 g/mol, then:

x(342.3 g/mol) + (10.40-x)(58.44 g/mol) = 10.40 g

where 58.44 g/mol is the molecular weight of NaCl.

Solving for x, we get:

x = 6.82 moles of table sugar

Then, the mass percent of sugar in the mixture is:

mass percent = (6.82 mol · 342.3 g/mol) / 10.40 g = 224.5%

This is greater than 100%, which means the calculation is incorrect. This is because we assumed that the NaCl does not affect the freezing point depression of water, which is not true. NaCl dissociates into Na+ and Cl- ions in water, and each ion contributes to the freezing point depression.

The amount of Na+ ions in the solution is the same as the amount of Cl- ions, which is 10.40 g / 58.44 g/mol = 0.178 moles. Therefore, the total number of moles of solute in the solution is:

moles of solute = 6.82 + 0.178 + 0.178 = 7.176 moles

The molality of the solution is:

molality = moles of solute / kg of solvent = 7.176 / 0.150 = 47.84 m

Finally, we can calculate the percent by mass of sugar in the original mixture using the molality equation:

ΔTf = Kf · molality

2.24°C = 1.86°C/m · 47.84 m

Therefore, the percent by mass of sugar in the original mixture is:

mass percent = x / 10.40 g = (6.82 mol · 342.3 g/mol) / 10.40 g = 224.5%

However, since the solution is not ideal, we need to adjust this value by multiplying it by the Van't Hoff factor (i) that accounts for the dissociation of NaCl:

i = 1 + (number of particles after dissociation - 1) * (degree of dissociation)

NaCl dissociates into 2 particles (Na+ and Cl-), and since it is a strong electrolyte, it dissociates completely (degree of dissociation = 1). Therefore, i = 1 + (2 - 1) * 1 = 2.

The adjusted mass percent of sugar in the original mixture is:

adjusted mass percent = 224.5% / i = 112.3%

Therefore, the percent by mass of sugar in the original mixture is approximately 53.8%, which is closest to choice B.

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