A solution is prepared by dissolving 15.0 g of nh3 in 250.0 g of water. the density of the resulting 7)solution is 0.974 g/ml. t

Question

drek231 [11]

3 years ago

7

A solution is prepared by dissolving 15.0 g of nh3 in 250.0 g of water. the density of the resulting 7)solution is 0.974 g/ml. t

he molality of nh3 in the solution is __________.

Chemistry

2 answers:

GaryK [48] · Answer 1 · 2022-01-25T04:02:21+03:00

0.0597
Mole fraction NH3 = moles NH3 / total moles in solution
moles NH3 = mass / molar mass
= 15.0 g / 17.034 g/mol
= 0.88059 moles
moles H2O = 250 g / 18.016 g/mol
= 13.8766 mol
total moles = 0.88059 + 13.8766
= 14.757 mol
fole fraction NH3 = 0.88059 / 14.757
= 0.0597

BARSIC [14] · Answer 2 · 2022-01-25T06:06:14+03:00

To overcome this problem, we first assume the volume is purely additive. The density of the mixture can then be calculated by adding up the mass fraction of each component divided by the density of each:

1 / ρ mixture = (x NH3 / ρ NH3) + (x H2O / ρ H2O) ---> 1

<h2>Further Explanation </h2><h3>Calculate the mass fraction of NH3: </h3>

x NH3 = 15 g / (15 g + 250 g)

x NH3 = 0.0566

<h3>Therefore the water mass fraction is: </h3>

x H2O = 1 - x NH3 = 1 - 0.0566

x H2O = 0.9434

<h3>Assuming that the density of water is 1 g / mL and replaces known values back to equation 1: </h3>

1 / 0.974 g / mL = [0.0566 / (ρ NH3)] + [0.9434 / (1 g / mL)]

NH3 = 0.680 g / mL

<h3>Given the density of NH3, we can now calculate the volume of NH3: </h3>

V NH3 = 15 g / 0.680 g / mL