Calculate the mass of ZnCl2 that was dissolved.

Question

A certain liquid X has a normal freezing point of −6.30°C and a freezing point depression constant K_f = 1.95·°C·kgmol^-1. A solution is prepared by dissolving some zinc chloride (ZnCl₂) in 500. g of X. This solution freezes at −7.1°C. Calculate the mass of ZnCl₂ that was dissolved.Round your answer to 1 significant digit.

Answer 1

∆T = imK

∆T = change in freezing point = 0.8º

i = van't Hoff factor = 3 (1Zn + 2Cl)

m = molality = moles ZnCl₂/kg solvent X = ?

K = freezing constant for solvent X = 1.95

Solve for m (molality):

m = ∆T / (i)(K) = 0.8 / (3)(1.95)

m = 0.1368 moles ZnCl₂/kg solvent X

Since we don't have 1 kg of solvent, but rather we have 500 g or 0.500 kg, we can solve for mass of ZnCl₂:

mass ZnCl₂ present = 0.1368 mols / kg x 0.500 kg = 0.0684 mols x 136.3 g ZnCl₂/mol = 9.32 g ZnCl₂

Rounding to 1 sig.fig. we have 9 g ZnCl₂