Model Question Paper — 1 — Question 20

(a) Molar mass of Mohr's salt [(NH₄)₂SO₄.FeSO₄.6H₂O]

= 2[14 + 4(1)] + 32 + 4(16) + 56 + 32 + 4(16) + 6[2(1) + 16]

= 2[18] + 32 + 64 + 56 + 32 + 64 + 6[18]

= 36 + 32 + 64 + 56 + 32 + 64 + 108

= 392 g

(ii) Molar mass of Mohr's salt [(NH₄)₂SO₄.FeSO₄.6H₂O] = 392 g

Mass of nitrogen in one mole of Mohr's salt = 2 x 14 = 28 g

392 g Mohr's salt has mass of nitrogen = 28 g

∴ 100 g Mohr's salt will have mass

= $\dfrac{28 \times 100 }{392}$

= 7.14%