## Chemistry (4th Edition)

$1420$ mL of $0.112$ ammonium sulfate.
1. Determine the molar mass of this compound ($NH_4^+$): 14.01* 1 + 1.008* 4 = 18.042g/mol 2. Calculate the number of moles: $mm(g/mol) = \frac{mass(g)}{n(mol)}$ $18.042 = \frac{5.75}{n(mol)}$ $n(mol) = \frac{5.75}{ 18.042}$ $n(mol) = 0.319$ ** Each ammonium sulfate $((NH_4)_2SO_4)$ contains 2 ammonium ions; therefore: $[NH_4^+] = [(NH_4)_2SO_4] * 2 = 0.112 M * 2= 0.224M$ 3. Find the volume: $Concentration(M) = \frac{n(mol)}{V(L)}$ $0.224 = \frac{0.319}{V(L)}$ $V(L) = \frac{0.319}{0.224}$ $V(L) = 1.42$ 4. Convert that number to mL: 1 L = $1000$ mL $1.42$ L = $1.42 \times 1000$ mL = $1420$ mL