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)
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)
Updated On: Jun 27, 2024
- 7 g
- 14 g
- 28 g
- 3.5 g
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The Correct Option is A
Solution and Explanation
${16.9 \; g \; AgNO_3}$ 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
$\therefore$ 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
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