Question

Determine the molality of 70.5% perchloric acid (HClO4)

Answer 1

Let mass of solution be 1 Kg = 1000 g

mass of HClO4 = 70.5 % of mass of solution

= 70.5*1000.0/100

= 705.0 g

mass of solvent = mass of solution - mass of solute

mass of solvent = 1000 g - 705.0 g

mass of solvent = 295.0 g

mass of solvent = 0.295 Kg

Molar mass of HClO4,

MM = 1*MM(H) + 1*MM(Cl) + 4*MM(O)

= 1*1.008 + 1*35.45 + 4*16.0

= 100.458 g/mol

mass(HClO4)= 705.0 g

use:

number of mol of HClO4,

n = mass of HClO4/molar mass of HClO4

=(7.05*10^2 g)/(1.005*10^2 g/mol)

= 7.018 mol

m(solvent)= 0.295 Kg

use:

Molality,

m = number of mol / mass of solvent in Kg

=(7.018 mol)/(0.295 Kg)

= 23.79 molal

Answer: 23.8 molal