Question

If the density of some lake water is 1.25 g mL⁻¹ and contains 92 g of Na⁺ ions per kg of water, calculate the molality of Na⁺ ions in the lake.

Answer 1

Given:

Mass of Na⁺ ions = 92 g

Mass of water (solvent) = 1 kg

Molar mass of Na⁺ = 23 g/mol

Formula:

Molality (m) = ${\displaystyle \frac{\text{Moles of solute}}{\text{Mass of solvent (kg)}}}$

Solution:

Number of moles present in 92 g of Na⁺ ions = ${\displaystyle \frac{\text{92 g}}{23 {\text{g mol}}^{-1}}}$ ...(as atomic mass of Na = 23)

= 4 mole

Molality (m) = ${\displaystyle \frac{\text{Moles of solute}}{\text{Mass of solvent (kg)}}}$

Therefore, molality of Na⁺ ions in the lake = ${\displaystyle \frac{4 \text{mole}}{1 \text{kg}}}$

= 4 mol/kg