Question

(10 points) A student gathers the following data for the freezing point depression of a pure solvent and solvent-solute solution. Based on this information, what is the molecular mass of the solute? Molality (m) Mass of solute (g) Mass of solvent (kg) 0.2214 0.632 0.00521

Answer 1

Molecular weight of the solute = 0.5479 kg/mol or 547.9g/mol

Step-by-step explanation:

Molarity,M is defined as the number of moles of solute dissolved in 1 liter of solution.

M = n/V

Where n is the number of moles

V is the volume of solution in liter

Molality,m is the number of moles of solute per kilogram of solvent.

molality = n/Ms

Where n is the number of solute

Ms is the kilogram of solvent in kg

In the information given,

Mass of solute = 0.632g

Molality, m = 0.2214

Mass of solvent, Ms = 0.00521kg

Using,

Molality, m = number of moles of solute/kilogram of solvent

Number of moles of solute = mass of solute in kg/molecular weight, M.W of solute in kg/mol

n = (0.632/1000)/M.W

Inputting all these values,

0.2214 = (6.32 x 10^-4/M.W)/0.00521

M.W = 6.32 x 10^-4/(0.00521 * 0.2214)

= 0.5479 kg/mol

= 547.9g/mol

Answer 2

548 g/mol

Step-by-step explanation:

The freezing point depression of a solvent occurs when a nonvolatile solute is added to it. Because of the interactions between solute-solvent, it is more difficult to break the bonds, so the phase change will need more energy, and the freezing point will drop, which is called cryoscopy.

The drop in temperature can be calculated by:

ΔT = Kf*W*i

Where Kf is the cryoscopy constant of the solvent, W is the molality, and i is the van't Hoff factor, which indicates the fraction of the solute that dissolves.

The molality represents how much moles (n) of the solute is presented in each kg of the solvent (m2), thus

W = n/m2

The number of moles is the mass of the solute (m1) in g, divided by the molar mass (M1) of it:

W = m1/(M1*m2)

So, by the data:

0.2214 = 0.632/(M1*0.00521)

0.00115M1 = 0.632

M1 = 548 g/mol