Question

What is the mass of the precipitate formed when 50 mL of 16.9% solution of $AgNO_3$ is mixed with 50 mL of 5.8% $NaCl$ solution ? (Ag = 107.8, N = 14, O = 16, Na = 23, Cl =35.5)

• A. 7 g
• B. 14 g
• C. 28 g
• D. 3.5 g

Answer 1

Answer: (A)

7 g

Answer 2

${16.9 \; g \; AgNO_3}$ is present in 100 mL solution.
$\therefore$ 8.45 g ${AgNO_3}$ is present in 50 mL solution
5.8 g NaCl is present in 100 mL solution
$\therefore$ 2.9 g NaCl is present in50 mL solution
\hspace20mm {AgNO_{3 } + NaCl -> AgCl + NaNO_{3}}
\hspace16mm{\frac{8.45}{170} mol} \hspace10mm \frac{2.9}{58.5}
\hspace8mm {= 0.049 mol} = {0.049 mol } \to 0 \hspace10mm 0
{after \\\ reaction } \hspace10mm 0 \hspace10mm 0 \to {0.049 mol} \; \; \; {0.049 mol}
mass of AgCl precipitated
$= 0.049 \times 143.5 g$
= 7 g Ag Cl