Calculate the number of NH4- ions in 3.63e4 mL of 6.5e-1 M ammonium sulfate.

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Ammonium ions have a charge of +1 when ionized.
(NH4)2SO4 --> 2NH4^+ + SO4^2-

3.63e4 mL is 36.3 L of solution.

36.3 L x (0.65 mol (NH4)2SO4 / 1 L) x (2 mol NH4+ / 1 mol (NH4)2SO4) = 110 mol NH4+ ..... round 111.692 mol to two significant digits

We can express the number of ions in moles, or we can calculate the number of ions using Avogadro's number.

111.692 mol NH4+ x (6.022x10^23 NH4+/ 1 mol NH4+) = 6.7x10^25 NH4+ ....... to two significant digits

Actually, 36.3 x 0.65 x 2 is 47.19.
So the answer should have been 47 mol NH4+, to two significant digits.

I must have hit the wrong calculator key and divided by 0.65. The odd thing is that my wrong answer matched your on-line answer.

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V of NH4+ ions = 3.63E4 ml
= 3.63 x 10^4 ml = 36.3 x 10^3 ml
= 36.3 L

Conc, M, of NH4+ ions = 6.5E-1 M
= 6.5 x 10^-1 M = 0.65 M
= 0.65 mol/L

Conc = Nos of moles / Vol
Nos of moles = Conc x Vol
= 36.3 L x 0.65 mol/L
= 23.595 moles

Using Avogadro's Nos,
1 mole of a substance = 6.023 x 10^23 particles/molecules/ions/atoms
Thus,
23.595 moles of NH4+ has;
= 23.595 x 6.023 x 10^23 ions
= 142.112685 x 10^23 ions
= 142.113 x 10^23 ions [ to 3 d.p.]
= 1.42113 x 10^25 ions

= 1.42113E25 ions