Question

An aqueous solution of copper(II) iodide has a concentration of 0.448 molal.

The percent by mass of copper(II) iodide in the solution is _______%

Answer 1

let mass of solvent be 1 Kg

m(solvent)= 1 Kg

number of mol,

n = Molality * mass of solvent in Kg

= (0.448 mol/Kg)*(1 Kg)

= 0.448 mol

Molar mass of CuI2,

MM = 1*MM(Cu) + 2*MM(I)

= 1*63.55 + 2*126.9

= 317.35 g/mol

mass of CuI2,

m = number of mol * molar mass

= 0.448 mol * 317.35 g/mol

= 142.2 g

SO,

mass of solution = mass of solvent + mass of CuI2

= 1000 g + 142.2 g

= 1142.2 g

mass % of CuI2 = mass of CuI2 * 100 / mass of solution

= 142.2 * 100 / 1142.2

= 12.4 %

Answer: 12.4 %