molality- chemistry

Question

A sodium hydroxide solution that contains 43.7 grams of NaOH per L of solution has a density of 1.08 g/mL. Calculate the molality of the NaOH in this solution.

Answer 1

Let’s assume we have 1 liter of solution.

1 L x1000 ml/L = 1000 ml

1000 ml x 1.08 g/ml = 1080 g of solution

How many grams of this is NaOH and how many g are water?

We have 43.7 g NaOH

We have 1080 g - 43.7 g = 1036.3 g of water

We are asked to find molality which is defined as moles/kg solvent

moles = 43.7 g x 1 mole/58.4 g= 0.748 mol NaCl

kg of solvent = 1036.3 g x1 kg/1000gm = 1.036 kg

molality (m) = 0.748 mol/1.036 kg = 0.722 m