Question

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?

Answer 1

0.768 kg

Step-by-step explanation:

Step 1: Given data

• Mass of ammonium nitrate (solute): 27.8 g

• Molality of the solution (m): 0.452 m

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`