What is the mass of the precipitate formed when 50 mL of 16.9% solution of AgNO3 is mixed with 50 mL of 5.8% NaCl solution? (Ag = 107.8, N = 14, O = 16, Na = 23, Cl = 35.5)
(a) 3.5 g
(b) 7 g
(c) 14 g
(d) 28 g
Ans. (b) 7g
Sol.
AgNO3 + NaCl —> NaNO3 + AgCl
w/v% of AgNO3 = 16.9%
thus 16.9g of AgNO3 — 100ml of solution
50ml of solution — 16.9/100 x 50 = 8.45g of AgNO3
w/v% of NaCl = 5.8%
thus 5.8g of NaCl— 100ml of solution
50ml of solution — 5.8/100 x 50 = 2.9g of NaCl
n = weight/molecular mass
no of moles of AgNO3 = 8.45/169.8 = 0.04
no of moles of NaCl = 2.9/58.5 = 0.04
both reactants completely get consumed
58.5g NaCl — 143.3g AgCl
2.9g NaCl — 143.3/58.5 x 2.9 = 7.1g AgCl