How many grams of water are needed to dissolve 27.8 g of ammonium nitrate NH 4NO 3 in order to prepare a 0.452 m solution?

Question

asked Oct 4, 2021 179k views

1 Answer

Will Ediger · Answer 1 · 2021-10-11T06:41:44+0000

Answer:

0.768 kg

Step-by-step explanation:

Step 1: Given data

Step 2: Calculate the moles corresponding to 27.8 g of ammonium nitrate

The molar mass of ammonium nitrate is 80.04 g/mol.

27.8 g * (1mol)/(80.04g) = 0.347mol

Step 3: Calculate the mass of water

The molality is equal to the moles of solute divided by the kilograms of solvent (water).

$m= (n(solute))/(m(solvent)) \\m(solvent) = (n(solute))/(m) = (0.347mol)/(0.452mol/kg) = 0.768 kg$