Question

The freezing point of water, H₂O, is 0.000 °C at 1 atmosphere. K_f(water) = 1.86 °C/m

In a laboratory experiment, students synthesized a new compound and found that when 14.38 grams of the compound were dissolved in 227.9 grams of water, the solution began to freeze at -0.343 °C. The compound was also found to be nonvolatile and a non-electrolyte.

What is the molecular weight they determined for this compound?

Answer 1

We have relationship between depression in F.P and molality of solution as, T _f = K _f m

Where, T _f is a depression in freezing point = F.P of solvent - F. P of solution

K _f is a freezing point constant of solvent & m is a molality of solution.

0.000 ⁰ C - ( - 0.343 ⁰  C ) = 1.86 ⁰ C / m molality of solution

0.343 ⁰  C= 1.86 ⁰ C / m molality of solution

molality of solution = 0.343 ⁰  C/ 1.86 ⁰ C / m

molality of solution =0.1844 mol / kg

We have, Molality of solution = No of moles of solute / mass of solvent in kg

No of moles of solute = Molality of solution mass of solvent in kg

No of moles of solute =0.1844 mol / kg 0.2279 kg

No of moles of solute = 0.04202 mol

We have, no. of moles = Mass / molar mass

Molar mass of solute = Mass / No. of moles of solute

Molar mass of solute =14.38 g / 0.04202 mol

Molar mass of solute = 342.2 g / mol

ANSWER : Molar mas of compound : 342 g / mol