Question

Determine the molality of ions in a solution formed by dissolving 0.187 moles of NaCl in 456 grams of water. The density of the solution is 1.44 g/mL

Answer 1

Answer:

Molality Ions Na⁺ = 0.410 m

Molality Ions Cl⁻ = 0.410 m

Total ion molality = 0.820 m

Explanation:

Molality (m) is defined as moles of solute that are dissolved in 1000g (1kg) of solvent. In this case, the molality of the ions is requested, so it should be borne in mind that when the NaCl salt dissolves in the water, it dissociates forming ions as follows:

NaCl ⇒Na⁺ (aq) + Cl⁻ (aq)

Therefore, for each mole of salt that dissolves, you will have 1 mole of Na⁺ and 1 mole of Cl⁻

Knowing the amount of initial solute moles and the mass of water, we proceed to perform the calculations for each ion:

Ions Na⁺

456 g of H₂O ____ 0.187 mol Na⁺

1000 g of H₂O ____ X = 0.410 m

Calculation: 1000 g x 0.187 mol / 456 g = 0.410 mol

Ions Cl⁻

456 g of H₂O ____ 0.187 mol Cl⁻

1000 g of H₂O ____ X = 0.410 m

Calculation: 1000 g x 0.187 mol / 456 g = 0.410 m

If both results are added, the total molality of the ions will be obeyed:

0.410 m + 0.410 m = 0.820 m

This means that for every kilogram of water, 0.820 moles of dissociated ions from NaCl are dissolved.