what is the total no of cations present in 196g of Mohr salt?

Question

answer should be written in NA (Avogadro number)

Answer 1

First you want to divide by the molar mass of Mohr's salt to convert over into moles.

196 g x (1 mol / 392.15 g) = 0.5 mol

meaning that the total number of Cations here should be

0.5 mol x 3 = 1.5 mol NA

Answer 2

The molecular mass of Mohr's salt is 392.14 g/mol. The NA is 2.00. Considering that Mohr salt has two ammonium and one ferrous (Fe(II)) cations, the total number of cations is NA = 6.00

Answer 3

Googling Mohr's Salt, we get the compound ammonium iron(II) sulfate, whose chemical formula is FeSO₄(NH₄)₂SO₄ its anhydrous state and FeSO₄(NH₄)₂SO₄ 6H₂O as a hexahydrate (complexed with water). It contains multiple cations, 2 NH₄⁺ and 1 Fe²⁺, for a total of 3 cations per formula unit.

The molar mass of Mohr’s salt (in its anhydrous state) is equal to 284.035 grams per mole. The hexahydrate of Mohr’s salt has a molar mass of 392.125 grams per mole. These are calculated by adding up the molar masses of each element. For example, for the anhydrous form

Fe + 2 S + 8 O + 2 N + 8 H = 55.845 + 2(32.06) + 8(15.999) + 2(14.007) + 8(1.008) = 284.035 g/mol

For the hydrate, simply add 12 H and 6 O for the extra 6 H2O

Fe + 2 S + 14 O + 2 N + 20 H = 55.845 + 2(32.06) + 14(15.999) + 2(14.007) + 20(1.008) = 392.125 g/mol

These are very close to the molar masses listed in byJ'us of 284.05 g/mol and 392.13 g/mol, respectively.

Assuming we have the hexahydrate form,

196 g FeSO₄(NH₄)₂SO₄ 6H₂O x 1 mol FeSO₄(NH₄)₂SO₄ 6H₂O / 284.035 g FeSO₄(NH₄)₂SO₄ 6H₂O x 3 mol cations/1 mol FeSO₄(NH₄)₂SO₄ 6H₂O x N _A cations / 1 mol cations

=2.07 N_A cations

A similar calculation for the anhydrous state will boil down to

196 g x 1 mol/392.125 g x 3 mol cations /1 mol x N_A /1 mol cations

= 1.50 N_A cations

I'm not sure if the questions wants you to multiply out Avogadro's number, but if so, take each answer x 6.022 x 10^23.