What is the mass of precipitate formed when $50 \mathrm{~mL}$ of $16.9 \%$ solution of $\mathrm{AgNO}_{3}$ is mixed with $50 \mathrm{~mL}$ of $5.8 \% \mathrm{NaCl}$ solution?
$(\mathrm{Ag}=107.8, \mathrm{~N}=14, \mathrm{O}=16, \mathrm{Na}=23, \mathrm{Cl}=35.5)$

Question

Aicharya · Accepted Answer

$50 \mathrm{~mL}$ of $16.9 \%$ solution of $\mathrm{AgNO}_{3}$
$\left(\frac{16.9}{100} \times 50ight)=8.45 \mathrm{~g}$ of $\mathrm{Ag} \mathrm{NO}_{3}$
$\mathrm{n}_{\mathrm{mole}}=\frac{8.45 \mathrm{~g}}{(107.8+14+16 \times 3) \mathrm{g} / \mathrm{mol}}=\left(\frac{8.45 \mathrm{~g}}{169.8 \mathrm{~g} / \mathrm{mol}}ight)$
$=0.0497 \mathrm{moles}$
$50 \mathrm{ml}$ of $5.8 \%$ solution of $\mathrm{NaCl}$ contain $\mathrm{NaCl}=\left(\frac{5.8}{100} \times 50ight)=2.9 \mathrm{~g}$
$\begin{aligned} \mathrm{n}_{\mathrm{NaCl}} &=\frac{2.9 \mathrm{~g}}{(23+35.5) \mathrm{g} / \mathrm{mol}}=0.0495 \mathrm{moles} \\ \mathrm{AgNO}_{3}+\mathrm{NaCl} & ightarrow \mathrm{AgCl} \downarrow+\mathrm{Na} \odot+\mathrm{Cl} \ominus \end{aligned}$
1 mole $\quad 1$ mole $\quad 1$ mole $\therefore 0.049$ mole $0.049$ mole $\quad 0.049$ mole of $\mathrm{AgCl}$
$\mathrm{n}=\frac{\mathrm{w}}{\mathrm{M}}$
$\mathrm{w} \quad=\left(\mathrm{n}_{\mathrm{AgC}}ight) \times$ Molecular Mass
$\quad=(0.049) \times(107.8+35.5)$
$\quad=7.02 \mathrm{~g}$ .

What is the mass of precipitate formed when $50 \mathrm{~mL}$ of $16.9 \%$ solution of $\mathrm{AgNO}_{3}$…

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