Question

If $x$ g weight of $AgCl$ will be precipitated when a solution containing $4.77$ g $NaCl$ is added to a solution of $5.77$ g of $AgN{O}_3$, then find the value of $x$.

• A. $143.5$
• B. $142.5$
• C. $147.5$
• D. None of the above

Answer 1

The correct option is A $143.5$

$AgN{O}_3+NaCl\to AgCl+NaN{O}_3$
Molecular weight of $AgN{O}_3=108+14+16\times 3=170$ g/mol
Molecular weight of $NaCl=23+35.5=48.5$ g/mol
$0.033$ moles of $AgCl$ is formed.
Weight of $AgCl$ $=0.033\times 143.5=4.7355$ g
Molecular weight of $AgCl=108+35.5=143.5$ g/mol