Asked by molality--->molarity

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.

Answered by bobpursley
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)

Answered by sara
how ccan i make nh4oh from nh3?
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