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(B) $1$. Calculate the moles of $AgNO_3$: The molar mass of $AgNO_3 = 107.8 + 14 + (3 	imes 16) = 169.8 \ g/mol$. Assuming the density of the solutio

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$7.17$

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(B) $1$. Calculate the moles of $AgNO_3$: The molar mass of $AgNO_3 = 107.8 + 14 + (3 	imes 16) = 169.8 \ g/mol$. Assuming the density of the solution is $1 \ g/mL$,mass of $50 \ mL$ solution = $50 \ g$. Moles of $AgNO_3 = (50 	imes 0.169) / 169.8 \approx 0.05 \ mol$.$2$. Calculate the moles of $NaCl$: The molar mass of $NaCl = 23 + 35.5 = 58.5 \ g/mol$. Moles of $NaCl = (50 	imes 0.058) / 58.5 \approx 0.05 \ mol$.$3$. The reaction is: $AgNO_3(aq) + NaCl(aq) ightarrow AgCl(s) + NaNO_3(aq)$.$4$. Since the stoichiometry is $1:1$,$0.05 \ mol$ of $AgNO_3$ reacts with $0.05 \ mol$ of $NaCl$ to produce $0.05 \ mol$ of $AgCl$ precipitate.$5$. Molar mass of $AgCl = 107.8 + 35.5 = 143.3 \ g/mol$.$6$. Mass of $AgCl = 0.05 \ mol 	imes 143.3 \ g/mol = 7.165 \ g \approx 7.17 \ g$.

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 (in $g$)? (Atomic masses: $Ag = 107.8, N = 14, O = 16, Na = 23, Cl = 35.5$)

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