Question

What is the mass of precipitate formed when 50 mL of 16.9% solution of $AgN{O}_3$ is mixed with 50 mL of 5.8% NaCl solution ?

(Ag = 107.8, N = 14, O = 16, Na = 23, Cl = 35.5)

• A. 18g
• B. 28g
• C. 7g
• D. 3.5g

Answer 1

Answer: (C)

7g

Moles of $AgN{O}_3=\frac{50\times 16.9}{100\times 169.8}=0.05$ moles
Moles of $NaCl=\frac{50\times 5.8}{100\times 58.5}=0.05$ moles
$AgN{o}_3+NaCl\to AgCl+NaN{O}_3$

Mass of AgCl $=$ mole x molar mass

$=0.05\times 143.5=7.16$ g

Hence option C is correct.