Question

What is the molality of a 2.04 M KBr aqueous solution?The density of the solution is 1.11 g/mL.

Answer 1

Let volume be 1 L

volume , V = 1 L

number of mol,

n = Molarity * Volume

= 2.04*1

= 2.04 mol

volume , V = 1 L

= 1*10^3 mL

density, d = 1.11 g/mL

mass = density * volume

= 1.11 g/mL *1*10^3 mL

= 1110 g

This is mass of solution

Molar mass of KBr,

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

= 1*39.1 + 1*79.9

= 119 g/mol

mass of KBr,

m = number of mol * molar mass

= 2.04 mol * 119 g/mol

= 2.428*10^2 g

This is mass of solute

mass of solvent = mass of solution - mass of solute

= 1110 - 242.76

= 867.24 g

= 0.86724 Kg

m(solvent)= 0.86724 Kg

Molality,

m = number of mol / mass of solvent in Kg

=(2.04 mol)/(0.86724 Kg)

= 2.352 molal

Answer: 2.35 molal