Question

An aqueous solution of potassium iodide has a concentration of 0.231 molal.

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

Answer 1

lets assume mass of solvent to be 1 Kg

m(solvent)= 1 Kg

number of mol,

n = Molality * mass of solvent in Kg

= (0.231 mol/Kg)*(1 Kg)

= 0.231 mol

Molar mass of KI,

MM = 1*MM(K) + 1*MM(I)

= 1*39.1 + 1*126.9

= 166 g/mol

mass of KI,

m = number of mol * molar mass

= 0.231 mol * 166 g/mol

= 38.35 g

so,

mass of solution = mass of solvent + mass of KI

= 1000 g + 38.35 g

= 1038.35 g

mass % of KI = mass of KI *100 / mass of solution

= 38.35 * 100 / 1038.35

= 3.69 %

Answer: 3.69 %