Question

Molar Mass by Freezing Point Depression I am doing a latenite lab on Molar Mass by Freezing Point Depression. I have asked this question on Course Hero and twice on Chegg and the tutors keep giving me different answers. Sample 2 has a van't Hoff factor of 2. It began freezing at -3.6. I added 2 g of the unknown to 10 mL of water. I was given that the Kf for water is 1.86.

Calculate the molar mass of FP sample 2

Answer 1

Given,

Van't Hoff factor = 2

Freezing point of solution = -3.6 ^oC

Mass of solute = 2g

The volume of solvent(Water) = 10 mL

We know, the formula,

T_f = T_solvent - T_solution ---------(1)

Also,

T_f = i x K_f x m --------(@)

Using formula (1) to calculate the freezing point depression value,

T_f = T_solvent - T_solution

T_f = 0 - (-3.6)

T_f = 3.6 ^oC

Now, using formula (2) to calculate the value of "m",

T_f = i x K_f x m

3.6 ^oC = 2 x 1.86 ^oC/m x molality

Molality(m) = 0.9677 m

Now, We know,

The density of water = 1 g/mL

Thus, converting the mass of solvent(water) to volume,

= 10 mL of water x ( 1g / 1 mL)

= 10 g of water

= 0.010 kg of water.

Now, We know,

Molality = Number of moles of solute / Kg of solvent

0.9677 = Number of moles of solute / 0.01 kg water

The number of moles of solute = 0.009677 moles

Now, the molar mass = grams / moles

molar mass = 2 g / 0.009677 moles

molar mass = 206.7 g /mol

Thus, the molar mass of FP sample 2 is 206.7 g/mol