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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

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Ans. (b) 7g

Sol.

AgNO3 + NaCl —> NaNO3 + AgCl

w/v% of AgNO3 = 16.9%

thus 16.9g of AgNO— 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 AgNO= 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

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