Question

A solution contains 3 g of mannose (C₆H₁₂O₆) in 50 mL acetic acid (C₂H₄O_2, density 1.05 g/mL) has a freezing point that is de by 1.23 celsius compared to pure acetic acid. Mannose doesn't dissociate in acetic acid. The freezing point for acetic is 16.6 celsius, while the boiling point is 118.1 celsius.

I. Determine the freezing point depression constant K_f for acetic acid.

II. If the boiling point elevation constant K_b for acetic acid is 3.07, determine the boiling point of acetic acid after the addition of 6 g of mannose is added to 50 mL acetic acid.

Answer 1

Formula, for non-dissociating or non-associating solutes:

and

where, f stands for freezing, b stands for boiling and C_m stands for molal concentration.

I.

Molar mass of mannose = 180 g.mol^-1 thus, moles of mannose in 3 g mannose = (3 / 180) mol = 0.0167 mol.

Amount of acetic acid = 50 mL = (50 x 1.05) g = 52.5 g

So, C_m = moles of solute in 1000 g of solvent = (0.0167 / 52.5) x 1000 = 0.3181 molal (m) or, 0.3181 mol.Kg^-1

So, from the formula,

NOTE: Change in tempreature is same in ^oC and K.

II.

6 g of mannose = 0.0333 mol mannose.

So, the corresponding molality of solution = (0.0333 / 52.5) x 1000 = 0.634

So,

So, boiling point of acetic acid will be (118.1 + 1.95) ^oC = 120.05 ^oC.