What is the mass of the precipitate formed when $100 \mathrm{~mL}$ of a $34.0\%$ (w/v) solution of $

Question

What is the mass of the precipitate formed when $100 \mathrm{~mL}$ of a $34.0\%$ (w/v) solution of $\mathrm{AgNO}_{3}$ is completely mixed with $100 \mathrm{~mL}$ of an $11.7\%$ (w/v) $\mathrm{NaCl}$ solution? (Molar mass: $\mathrm{Ag}=108, \mathrm{N}=14, \mathrm{O}=16, \mathrm{Na}=23, \mathrm{Cl}=35.5$)

Professor PC Thomas Classes · Answer

(a) : $34.0\% 	ext{ (w/v) } \mathrm{AgNO}_3$ means $34.0 \mathrm{~g}$ of $\mathrm{AgNO}_3$ is present in $100 \mathrm{~mL}$ of solution. The molar mass of $\mathrm{AgNO}_3$ is $108 + 14 + 48 = 170 \mathrm{~g/mol}$. Moles of $\mathrm{AgNO}_3 = \frac{34.0}{170} = 0.2 \mathrm{~mol}$. Similarly, $11.7\% 	ext{ (w/v) } \mathrm{NaCl}$ means $11.7 \mathrm{~g}$ of $\mathrm{NaCl}$ is present in $100 \mathrm{~mL}$. Molar mass of $\mathrm{NaCl} = 58.5 \mathrm{~g/mol}$. Moles of $\mathrm{NaCl} = \frac{11.7}{58.5} = 0.2 \mathrm{~mol}$. The reaction is $\mathrm{AgNO}_3 + \mathrm{NaCl} ightarrow \mathrm{AgCl} \downarrow + \mathrm{NaNO}_3$. Since both reactants are present in $0.2 \mathrm{~mol}$, exactly $0.2 \mathrm{~mol}$ of $\mathrm{AgCl}$ will precipitate. Mass of $\mathrm{AgCl} = 0.2 	imes (108 + 35.5) = 0.2 	imes 143.5 = 28.7 \mathrm{~g}$.