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

Answer:

0.768 kg

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.

$27.8 g \times \frac{1mol}{80.04g} = 0.347mol$

Step 3: Calculate the mass of water

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

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