Answer
$1420$ mL of $0.112$ ammonium sulfate.
Work Step by Step
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
