Question

A beaker contains 450 g of water (H2O). If 9.2 g of calcium fluoride (CaF2) is added, what is the molality concentration of the solution?

Answer 1

Molality = number of mole of solute in 1kg= 1000g solvent.

Solute is CaF2.
Molar mass (CaF2) = M(Ca) +2M(F) = 40.1 + 2*19.0 = 78.1 g/mol

9.2 g CaF2 * 1 mol CaF2/78.1 g CaF2 = 0.12 mol CaF2

Solvent is water 450 g.

0.12 mol CaCl2 -------450 g water
x mol CaCl2 --------- 1000 g water

x=0.12*1000/450 = 0.27

Molality is 0.27 (mol CaF2 in 1 kg water).