Question

An aqueous solution of sodium iodide has a concentration of 7.13×10-2 molal.

The percent by mass of sodium iodide in the solution is ________%.

Answer 1

The concentration of the aqueous solution of sodium iodide is 7.13*10-2 molal. We define concentration in molality as

Concentration in molality = (moles of solute)/(kg of solvent)

Let us assume that we have 1 kg of solvent. Therefore,

7.13*10-2 m = moles of NaI/(1 kg water)

===> moles of NaI = (7.13*10-2 m)*(1 kg water) = 7.13*10-2.

Molar mass of NaI = (1*22.989 + 1*126.904) g/mol = 149.893 g/mol.

Mass of NaI taken in solution = (7.13*10-2 mole)*(149.893 g/mol) = 10.6874 g.

Therefore, 1 kg water contains 10.6874 g NaI.

Therefore, 100 g water will contain (10.6874 g)*(100 g/1000 g) = 1.06874 g NaI (1 kg = 1000 g).

Total mass of the solution = (1.06874 + 100) g = 101.06874 g.

Mass% NaI in the solution = (1.06784 g)/(101.06784 g)*100 = 1.0574% ≈ 1.06% (ans).