Question

A solution is prepared by mixing 0.3355 moles of NaNO3 (84.995 g mol-1) with 235.0 g of water (18.015 g mol-1). Its density is 1.0733 g mL-1. What is the molality of the solution?

Answer 1

Molality = 1.428m

Step-by-step explanation:

Molality, m, is an unit of concentration defined as the ratio between moles of solute and kg of solvent:

Molality: Moles solute / kg solvent.

In the problem, water is the solvent (Is the compound in the higher quantity) whereas NaNO₃ is the solute.

Moles of solute: 0.3355moles NaNO₃.

Kg solvent:

`235.0g(1kg)/(1000g) = 0.2350kg`

Thus, molality of the solution is:

Molality = 0.3355 moles NaNO₃ / 0.2350kg

Molality = 1.428m