To solve this problem, first we assume the volume is
purely additive. The density of the mixture can then be calculated by the
summation of mass fraction of each component divided by its individual density:
1 / ρ
mixture = (x NH3 / ρ
NH3) + (x H2O / ρ<span>
H2O) ---> 1</span>
Calculating for mass fraction of NH3:
x NH3 = 15 g / (15 g + 250 g)
x NH3 = 0.0566
Therefore the mass fraction of water is:
x H2O = 1 – x NH3 = 1 – 0.0566
x H2O = 0.9434
Assuming that the density of water is 1 g / mL and substituting
the known values back to equation 1:
1 / 0.974 g / mL = [0.0566 / (ρ NH3)]
+ [0.9434 / (1 g / mL)]
ρ NH3
= 0.680 g / mL
Given the density of NH3, now we can calculate for the
volume of NH3:
V NH3 = 15 g / 0.680 g / mL
V NH3 = 22.07 mL
The number of moles NH3 is: (molar mass NH3 is 17.03 g /
mol)
n NH3 = 15 g / 17.03 g / mol
n NH3 = 0.881 mol
Therefore the molarity of NH3 in the solution is:
<span>Molarity = 0.881 mol / [(22.07 mL + 250 mL) * (1L / 1000 mL)</span>
<span>M = 3.238 mol/L = 3.24 M</span>
