Respuesta :
Answer:
0.00725 M
Explanation:
Considering for [tex]NH_4NO_3[/tex]
Mass = 5.66 g
Molar mass of [tex]NH_4NO_3[/tex] = 80.043 g/mol
Moles = Mass taken / Molar mass
So,
Moles = 5.66 / 80.043 moles = 0.0707 moles
[tex]NH_4NO_3[/tex] will dissociate as:
[tex]NH_4NO_3\rightarrow NH_4^++NO_3^-[/tex]
Thus 1 mole of [tex]NH_4NO_3[/tex] yields 1 mole of ammonium ions. So,
Ammonium ions furnished by [tex]NH_4NO_3[/tex] = 1 × 0.0707 moles = 0.0707 moles
Considering for [tex](NH_4)_3PO_4[/tex]
Mass = 4.42 g
Molar mass of [tex](NH_4)_3PO_4[/tex] = 149.09 g/mol
Moles = Mass taken / Molar mass
So,
Moles = 4.42 / 149.09 moles = 0.0296 moles
[tex](NH_4)_3PO_4[/tex] will dissociate as:
[tex](NH_4)_3PO_4\rightarrow 3NH_4^++PO_4^{3-}[/tex]
Thus 1 mole of [tex]NH_4NO_3[/tex] yields 3 moles of ammonium ions. So,
Ammonium ions furnished by [tex](NH_4)_3PO_4[/tex] = 3 × 0.0296 moles = 0.0888 moles
Total moles of the ammonium ions = 0.0707 + 0.0888 moles = 0.1595 moles
Given that:
Volume = 22.0 L
So, Molarity of the [tex]NH_4^+[/tex] is:
Molarity = Moles / Volume = 0.1595 / 22 M = 0.00725 M
