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.

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)

how ccan i make nh4oh from nh3?

To find the molarity (M) of a solution, you need to determine the number of moles of solute (ammonia, NH3) present in a given volume of solvent (water, H2O). Here's how you can calculate the molarity:

1. Start by finding the mass of the 0.87 m aqueous solution. Multiply the density (0.823 g/mL) by the volume of the solution. Let's assume the volume is 1 L (since 1 mL = 1 g). Multiply 0.823 g/mL by 1000 mL to get the mass of the solution in grams: 0.823 g/mL * 1000 mL = 823 g.

2. Calculate the mass of NH3 in the solution. Since the solution is 0.87 m, 1 L of the solution contains 0.87 moles of NH3. The molar mass of NH3 is 17.034 g/mol. Multiply the number of moles (0.87 mol) by the molar mass (17.034 g/mol) to get the mass of NH3 in grams: 0.87 mol * 17.034 g/mol = 14.807 g.

3. Find the number of moles of NH3. Divide the mass of NH3 by its molar mass: 14.807 g / 17.034 g/mol ≈ 0.868 mol.

4. Finally, calculate the molarity (M) by dividing the moles of NH3 by the volume of the solution in liters: 0.868 mol / 1 L = 0.868 M. Rounding to the appropriate number of significant figures, the molarity of the 0.87 m aqueous solution of ammonia is approximately 0.87 M.

Note that in the calculations, you assumed that 1 L of the solution weighs 823 g (using the density). However, in a true aqueous solution, the weight of the water is not equal to the total weight of the solution. That's why you haven't been able to separate the moles of NH3 from the kg of H2O or isolate the number of liters of H2O. In this case, it is more appropriate to calculate the mass of the solution and then determine the mass of NH3 by considering the solution's molarity.