Question

A 1.24M solution of sodium propionate (NaCH3CH2COO) with a density of 1.15 g/mL has an osmotic pressure of 48.5 atm at 16oC. Determine the van’t Hoff factor and the freezing point of the solution. Explain why the van’t Hoff factor is smaller than for an ideal solution.

Answer 1

Given info : A 1.24M solution of sodium propionate (NaCH3CH2COO) with a density of 1.15 g/mL has an osmotic pressure of 48.5 atm at 16°C.

To find : The Van't Hoff factor and the freezing point of the solution.

solution : using formula, π = iCRT

where π is osmotic pressure.

i is Van’t Hoff factor.

R is universal gas constant

and T is temperature in kelvin

here, π = 48.5 atm, C = 1.24 M , R = 0.082 atm.L/mol/K , T = 16 + 273 = 289 K

now, 48.5 = i × 1.24 × 0.082 × 289

⇒i = (48.5)/(1.24 × 0.082 × 289)

= i = 1.65

now using formula, ∆T_f = K_f × m

here we need K_f to find freezing point of the solution.

well, molality of the solution, m :

molarity is 1.24 M means, 1.24 moles of solute dissolved in 1000 mol of solution.

so, mass of solution = density of solution × volume of solution = 1.15 g/ml × 1000 ml = 1150 g

mass of solute = no of moles × molecular mass

= 1.24 × 96 g/mol

= 119.04 g ≈ 119 g

now mass of solvent = mass of solution - mass of solute

= 1150 - 119 = 1031g

now molality = 1.24 mol/1.031 kg = 1.2 molal

Van’t Hoff is small than for an ideal solution. it is because Van't Hoff factor depends on no of ions formed after dissociation. for ideal solution, it just dissociates and forms different ions but in real there are some ions which associates too, that's the reason no of total ions after dissociation in real solution is less than that for ideal solution.

That's why the Van't Hoff factor is less than for an ideal solution.