Question

What is the mass of precipitate formed when $50\text{mL}$ of $17\%\left(w/V\right)$ solution of $AgN{O}_3$ is mixed with $50\text{mL}$ of $5.85\%\left(w/V\right)$ $NaCl$ solution?
$(Ag=108,N=14,O=16,Na=23,Cl=35.5)$

• A. $3.5\text{g}$
• B. $7\text{g}$
• C. $14\text{g}$
• D. $28\text{g}$

Answer 1

The correct option is B $7\text{g}$

$17\%\left(w/V\right)$ solution of $AgN{O}_3$ means $17\text{g}$ of $AgN{O}_3$ in $100\text{mL}$ of solution.
Therefore $8.5\text{g}$ of $AgN{O}_3$ is present in $50\text{mL}$ solution.
Similarly, $5.85\text{g}$ of $NaCl$ is present in $100\text{mL}$ solution.
Therefore $2.925\text{g}$ of $NaCl$ is present in $50\text{mL}$ solution.
The reaction can be represented as
$$\begin{array}{ccccc}& AgN{O}_3+& NaCl\to & AgCl\downarrow +& NaN{O}_3\\ \text{Initial mole}& 8.5/170& 2.925/58.5& & \\ =& 0.05& 0.05& & \\ \text{Final mole}& 0& 0& 0.05& 0.05\end{array}$$
$\therefore$Mass of $AgCl$ precipitated $=0.05\times 143.5$
$=7.175\approx 7\text{g}$