Question

Question 2

An aqueous solution of sodium carbonate, Na₂CO₃ (MM = 105.99 g/mol) has a density of 1.43 g/mL and a molality of 14.2 m (or mol/kg). Calculate (i) the molarity of the solution, and (ii) the mole fraction of Na₂CO₃.

Answer 1

a)

assume 1 kg of solvent --> 1000 g of solvent

so we have 14.2 mol of solute = mass = mol*MW = 105.99*14.2 = 1505.058

total mass of solution = 1000+1505.058 = 2505.058 g of solution

volume = mass/D = 2505.058 / 1.43 = 1751.7888 mL

so

M = mol/V = 14.2 / (1751.7888*10^-3) = 8.105 molar

b)

mol fraction = mol of Na2CO3 / total mol

mol of Na2CO3 = 14.2 mol

mol of solvnt = 1000/18 = 55.555

mol fraction = mol of Na2CO3 / total mol = (14.2)/(14.2+55.555) = 0.2035