Question

What is the mass of the precipitate formed when $50$mL of $16.9$% solution of $AgN{O}_3$ 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

Answer 1

The correct option is A $7$ g

$50mL16.9\%AgN{O}_3=\frac{50\times 16.9}{100\times 169}=0.5molAgN{O}_3$

$50mL15.8\%NaCl=\frac{50\times 5.8}{100\times 5.8}=0.05molNaCl.$

$AgN{O}_3\underset{aq}{+}NaCl\underset{aq}{\to }AgCl\downarrow +NaN{O}_3\left(aq\right)$

$0.05$ $0.05$ - -

- - $0.05$ $0.05$

molecular mass of $AgCl$=143.3

mass of ppt$\left(AgCl\right)$ =$0.05\times 143.3$

$=7.1gm$

Hence, option $A$ is correct.