Question

The density of 3M solution of NaCl is 1.25g/mL. Calculate the molality of the solution. (Given: Atomic masses: Na=23u , Cl=35.5u)

Answer 1

First, let's calculate the mass of NaCl in 1L of the solution.

Molarity (M) = moles of solute/Liter of solution.

So, moles of NaCl in 1L of 3M solution = 3 moles.

The molar mass of NaCl = 23 (Na) + 35.5 (Cl) = 58.5 g/mol.

So, the mass of NaCl in 1L of the solution = 3 moles * 58.5 g/mol = 175.5 g.

The density of the solution is 1.25 g/mL, so the mass of 1L (or 1000 mL) of the solution = 1.25 g/mL * 1000 mL = 1250 g.

The mass of the solvent (water) = mass of the solution - mass of the solute = 1250 g - 175.5 g = 1074.5 g = 1.0745 kg.

Molality (m) = moles of solute/kg of solvent.

So, the molality of the solution = 3 moles/1.0745 kg = 2.79 mol/kg.