Question

A certain liquid X has a normal freezing point of 7.80 °C and a freezing point depression constant K,=3.03 °C kg mol. Calculate the freezing point of a solution made of 15.6g of urea (CH N,0) dissolved in 200. g of X. Be sure your answer is rounded to the correct number of significant digits. O°C

Answer 1

We have relationship between depression in F.P and molality of solution as, \(\Delta T_{f}=i \times K_{f} \times m\)

Where,i is a van't Hoff factor, \(\Delta T_{f}\) is a depression in freezing point = F.P of solvent - F. P of solution

\(\mathrm{K}_{\mathrm{f}}\) is a freezing point constant of solvent \(\& \mathrm{~m}\) is a molality of solution.

Urea is non electrolyte, hence \(\mathrm{i}=1\).

Calculation of molality

We know that, Molality = No. of moles of solute / mass of solvent in \(\mathrm{kg}\)

We have, No. of moles = Mass / Molar mass

Molar mass of Urea \((\mathrm{CH} 4 \mathrm{~N} 2 \mathrm{O})=(12.01)+(4 \times 1.0079)+(2 \times 14.0067)+16.00=60.06 \mathrm{~g} / \mathrm{mol}\)

\(\therefore\) No. of moles of urea \(=15.6 \mathrm{~g} /(60.06 \mathrm{~g} / \mathrm{mol})=0.2597 \mathrm{~mol}\)

We have, no. of moles of urea \(=0.2597 \mathrm{~mol}\), mass of solvent \((\mathrm{X})=200.0 \mathrm{~g}=0.2000 \mathrm{~kg}\)

\(\therefore\) Molality of urea solution \(=0.2597 \mathrm{~mol} / 0.2000 \mathrm{~kg}=1.298 \mathrm{~m}\)

We have relation, F.P of solvent \(-\mathrm{F}\). \(\mathrm{P}\) of solution \(=\mathrm{i} \times \mathrm{K}_{\mathrm{f}} \mathrm{X} \mathrm{m}\)

\(\therefore 7.80^{0} \mathrm{C}-\mathrm{F} . \mathrm{P}\) of urea solution \(=1 \times 3.03^{\circ} \mathrm{C} / \mathrm{m} \times 1.298 \mathrm{~m}\)

\(7.80^{0} \mathrm{C}-\mathrm{F} . \mathrm{P}\) of urea solution \(=3.9329^{\circ} \mathrm{C}\)

F. \(P\) of urea solution \(=7.80^{0} \mathrm{C}-3.9329^{\circ} \mathrm{C}\)

F. P of urea solution \(=3.87^{\circ} \mathrm{C}\)

ANSWER : F. P of urea solution \(=3.87^{\circ} \mathrm{C}\)