Question

An aqueous solution is 5.50% by mass ammonia, NH₃, and has a density of 0.975 g/mL.

The molality of ammonia in the solution is _______ m

Answer 1

consider 1 Kg of ammonia solution

mass of solution = 1 Kg = 1000 g

mass of NH3 = 5.50 % of 1000 g

= 5.50*1000/100

= 55.0 g

mass of solvent= remaining mass

= 1000 g - 55.0 g

= 94.5 g

Molar mass of NH3,

MM = 1*MM(N) + 3*MM(H)

= 1*14.01 + 3*1.008

= 17.034 g/mol

mass(NH3)= 55.0 g

number of mol of NH3,

n = mass of NH3/molar mass of NH3

=(55.0 g)/(17.034 g/mol)

= 3.229 mol

m(solvent)= 94.5 g

= 0.0945 Kg

Molality,

m = number of mol / mass of solvent in Kg

=(3.229 mol)/(0.0945 Kg)

= 34.2 molal

Answer: 34.2 m