Question

What is the molality of a semimolar NaCl solution if the density of the solution is 1.16 g cm⁻³?

Answer 1

Given: Molarity (M) = 0.5 mol/L (semimolar)

Density of solution = 1.16 g/cm³ = 1160 g/L

Molar mass of NaCl = 58.5 g/mol

We are to calculate molality (m).

For 1 L of solution:

Volume = 1 L

Mass of solution = 1160 g

Moles of NaCl = 0.5 mol

Mass of NaCl = 0.5 mol × 58.5 g/mol = 29.25 g

Mass of solvent (water) = 1160 g − 29.25 g = 1130.75 g = 1.13075 kg

Molality (m) = `"Moles of solute"/"Mass of solvent in kg"`

Molality (m) = `0.5/1.13075`

Molality (m) = 0.442 mol/kg

∴ Molality of a semimolar NaCl solution is 0.44 mol/kg.