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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
Detailed Solution
50 mL of 16.9% $AgNO_3$ contains 8.45 g = $\frac{8.45}{170} = 0.049$ mol
50 mL of 5.8% NaCl contains 2.9 g = $\frac{2.9}{58.5} = 0.049$ mol
$AgNO_3 + NaCl \rightarrow AgCl + NaNO_3$, so 0.049 mol AgCl forms.
Mass of AgCl = $0.049\times 143.5 \approx 7$ g
50 mL of 5.8% NaCl contains 2.9 g = $\frac{2.9}{58.5} = 0.049$ mol
$AgNO_3 + NaCl \rightarrow AgCl + NaNO_3$, so 0.049 mol AgCl forms.
Mass of AgCl = $0.049\times 143.5 \approx 7$ g