Question

Calculate the molarity of the resulting solution prepared by diluting 25.0 mL of 18.0% ammonium chloride, NH₄Cl, (density = 1.05 g/mL) to a final volume of 80.0 mL.

Answer 1

1st find the initial concentration of NH4Cl

Let volume of solution be 1 L

volume , V = 1 L

= 1*10^3 mL

density, d = 1.05 g/mL

use:

mass = density * volume

= 1.05 g/mL *1*10^3 mL

= 1.05*10^3 g

This is mass of solution

mass of NH4Cl = 18.0 % of mass of solution

= 18.0*1050.0/100

= 189.0 g

Molar mass of NH4Cl,

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

= 1*14.01 + 4*1.008 + 1*35.45

= 53.492 g/mol

mass(NH4Cl)= 189.0 g

use:

number of mol of NH4Cl,

n = mass of NH4Cl/molar mass of NH4Cl

=(1.89*10^2 g)/(53.49 g/mol)

= 3.533 mol

volume , V = 1 L

use:

Molarity,

M = number of mol / volume in L

= 3.533/1

= 3.533 M

use dilution formula

M1*V1 = M2*V2

1---> is for stock solution

2---> is for diluted solution

Given:

M1 = 3.533 M

V1 = 25 mL

V2 = 80 mL

use:

M1*V1 = M2*V2

M2 = (M1*V1)/V2

M2 = (3.533*25)/80

M2 = 1.1041 M

Answer: 1.10 M