Question

when 45.8 g of alanine (C3H7NO2) are dissolved in 1200. g of a certain mystery liquid X, the freezing point of the solution is 4.10 ℃ less than the freezing point of pure X. Calculate the mass of sodium chloride that must be dissolved in the same mass of X to produce the same depression in freezing point. The van't Hoff factor i-1.82 for sodium chloride in X. Be sure your answer has a unit symbol, if necessary, and is rounded to the correct number of significant digits x n?

Answer 1

16.5 g

Step-by-step explanation:

When a nonvolatile compound is dissolved in a pure solvent, the freezing point of the solvent is reduced, because the interaction solvent-solute requires more energy to be joined, and so, it freezes. This property is called cryoscopy, and the temperature change (ΔT) can be calculated by:

ΔT = Kc*W*i

Where Kc is the cryoscopy constant of the solute X, W is the molality of the solution, and i is the van't Hoff factor, which determines the percent of the solute that is dissolved. For organic molecules, such as alanine, i = 1.

The molality is the number of moles of the solute divided by the mass (in kg) of the solvent (1200 g = 1.2 kg). The molar mass of alanine is 89.09 g/mol, and the number of moles of it is the mass divided by the molar mass:

n = 45.8/89.09

n = 0.5141 mol

W = 0.5141/1.2 = 0.4284 mol/kg

So, Kc of X is:

4.10 = Kc*0.4284*1

Kc = 9.57 °C.kg/mol

So, if now sodium chloride is added to X, and the variation temperature is the same, and i = 1.82:

4.10 = 9.57*W*1.82

W = 0.2354 mol/kg

The number of moles of the solute is then:

W = n/1.2

0.2354 = n/1.2

n = 0.2825 mol

The molar mass of sodium chloride is 58.44 g/mol, thus the mass is the molar mass multiplied by the number of moles:

m = 58.44*0.2825

m = 16.5 g