Solve the below problems relating to the molality of NaNO3 in aqueous solutions. Make use of the Periodic Table on page 2 to calculate the molar masses.

(i) A solution is prepared using 17 g of water and 1 g of NaNO3. What is the molality of the NaNO3 solution?

(ii) In order to prepare a solution of 2 molal NaNO3 using 30 g of water,

what mass of NaNO3 is required?

i - Molality → 0.69 m

ii- 5.1 g of NaNO₃ are required

Explanation:

Our solute is the NaNO₃.

Molality is a sort of concentration, that indicates the moles of solute in 1kg of solvent. (mol/kg)

Molality → moles of solute / kg of solvent

We convert the mass of solvent, from g to kg

17 g . 1kg / 1000 g = 0.017kg

We determine the moles of NaNO₃ → 1 g / 85 g/mol = 0.0117 moles

i - Molality → 0.0117 mol / 0.017 kg = 0.69 m

ii- Molality is 2 m

It means that 2 moles of solute is contained in 1kg of solvent

We convert the mass of solvent from g to kg

30 g . 1kg / 1000 g = 0.030 kg

Molality = mol / mass of solvent (kg) → mol = Molality . mass of solvent(kg)

mol = 2 m . 0.030kg → 0.060 moles

We convert the moles to mass → 0.060 mol . 85 g / 1 mol = 5.1 g of NaNO₃

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