Question

Solve in complete solution and Explain briefly:

An organic compound is known to be non-volatile and non-electrolyte. A 0.5-g sample is dissolved in water and diluted to 160 mL. The osmotic pressure is measured as 0.05 atm at 30°C. The approximate mass number for this compound is 1400 g/mol.

From the data provided in this problem and knowing that the density of the solution is 1.00 g/mL.

1. Calculate the freezing point of the solutions.

2. Determine if the freezing point change would be a good way to determine the molecular mass of the compound.

3. Would the boiling point change be a better determining factor than the freezing point change?

Answer 1

$\Pi = icRT$

$i = Π/cRT\\ T = 30°C = 273 + 30°C = 303K\\ Π = 0.05atm = 0.05 × 101325 = 5066.25$

number of moles $(n) = 0.5/1400 =$

$0.000357$

concentration (c) = $\dfrac{0.000357}{160} × 1000$

$= 0.00223M = 2.23 mol/m³$

$i= \dfrac{5066.25}{ 2.23×8.314×303} = 0.9$

1. $∆T = iK_fm$

$K_f= 1.86K·kg/mol$

$i =0.9$

molality (m) = $\dfrac{0.000357}{0.160}= 0.00223$

$∆T = 0.9 × 1.86× 0.00223=$

$0.00373K$

2. Yes

3. No, because the magnitude of the freezing point depression is larger than the boiling point elevation for the same solvent and the same concentration of a solute.