Question

Data were collected to determine the molar mass of a nonvolatile solid solute dissolved in cyclohexane. Complete the
table for the analysis. Record calculated values with the correct number of significant figures. Show detailed
calculations to the right of the table.

Volume of cyclohexane (mL): 15.5
Mass of cyclohexane (kg): _________
Mass of added solute (g): 0.1660
Kf for cyclohexane (˚C/m): 20.0
Freezing point change, ΔTf (˚C): 2.20
Moles of solute (mol): _________
Molar mass of solute (g/mol):__________

Answer 1

We have the density of cyclohexane = 0.779 g/mL

Given that the volume of cyclohexane given is 15.5 mL

The weight of cyclohexane = 15.5 mL x 0.779 g/mL = 12.0745 grams = 0.0120745 kg

T_f = K_f m

K_f is the freezing point depression constant = 20 ^oC/m

T_f = depression in freezing point = 2.20 ^oC

'm' is the molality of the solution = number of moles of solutes/ Kg of the solvent

This can be written as

Molality = Weight of the solute / Molar mass of the solute x Kg of the solvent

Given weight of the solute is 0.1660 grams

Now, we can write the formula of depression in freezing point as

T_f = (K_f x Weight of the Solute)/Molar mass of Solute x Kg of the solvent

Molar Mass of Solute = (K_f x Weight of the Solute)/T_f x Kg of the solvent

Therefore, the Molar mass of solute = (20 ^oC/m x 0.1660 grams)/ 2.20 ^oC x 0.0120745 kg = 3.32 / 0.0265639 g/mol = 124.98 g/mol

Here the units of K_f are ^oC/molal solution and hence we cancelled out kg in the denominator while giving units to molar mass

Thus the required molar mass of the solute is 124.98 g/mol

Number of moles of solute = Weight of the solute/Molar mass of the solute = 0.1660 grams/124.98 g/mol = 0.00133 moles = 1.33 x 10^-3 moles.