Question

What is the molality (m) of 6.84% HCl by weight of an aqueous hydrochloric acid solution?

Answer 1

Let mass of solution be 1 Kg = 1000 g

mass of HCl = 6.84 % of mass of solution

= 6.84*1000.0/100

= 68.4 g

mass of solvent = mass of solution - mass of solute

mass of solvent = 1000 g - 68.4 g

mass of solvent = 931.6 g

mass of solvent = 0.9316 Kg

Molar mass of HCl,

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

= 1*1.008 + 1*35.45

= 36.458 g/mol

mass(HCl)= 68.4 g

use:

number of mol of HCl,

n = mass of HCl/molar mass of HCl

=(68.4 g)/(36.46 g/mol)

= 1.876 mol

m(solvent)= 0.9316 Kg

use:

Molality,

m = number of mol / mass of solvent in Kg

=(1.876 mol)/(0.9316 Kg)

= 2.014 molal

Answer: 2.01 molal