Nuts to what you did.
You have.87m, which assume 1000 ml of solution, or using density,
density= mass/v=.823g/ml=mass/1000ml
masssolution= 823 g, but now, one has to get the mass of the solute.
Now, some algebra.
m=molessolute/kgsolvent
.87*(kgsolvent)=masssolute/17
17 is mole mass of NH3...
well, kgsolvent is 823-mass solute, put that in, and solve for mass solute.
M= masssolute/(17*literssolution)
M= masssolute/(17*1.0)
what is the Molarity (M) of a 0.87m aqueous solution of ammonia, NH3?
The density of the solution is 0.823 g/mL.
Answer: 0.71 M
So I have:
0.823 g H2O+NH3/1 ml H2O+NH3
17.034 g NH3/1 mol
18.016 g H2O/1 mol
0.87 mol NH3/1 kg H2O
I've tried this several ways and I can't seem to separate to moles of NH3 from kg of H20 or isolate the number of Liters of H20.
I know (through calculations) that there are also 823 g NH3+H2O/L NH3+H2O
and 23.48 mol NH3+H2O/L NH3+H2O.
Any help on where to go would be appreciated.
2 answers
how ccan i make nh4oh from nh3?