Question

An aqueous NaCl solution is made using 116 g of NaCl diluted to a total solution volume of 1.30 L . Part A Calculate the molarity of the solution. Express your answer using three significant figures. Part B Calculate the molality of the solution. (Assume a density of 1.08 g/mL for the solution.) Express your answer using two significant figures.

Answer 1

  Part A:

MV = mass / molar mass

(x) (1.30 L) = 116 g / 58.443 g/mol

x = 1.9848/1.30 M = 1.53M; (3 significant figures)

Part B:

molality is moles solute per kg of solvent. I will use 1.9848 mol

Let us assume 1000 mL of the solution is present. This tells us that 1.9848 mol of the solute is present (that's the 116 g of NaCl).

1.08 g/mL times 1000 mL = 1080 g <--- this is the total mass of the 1000 mL solution

1080 g minus 116 g = 964 g <--- the mass of water in the 1000 mL of solution

964 g = 0.964 kg

1.9848 mol / 0.964 kg =2.1 m (using 2 significant figures)