Question

An excess of $AgN{O}_3$ is added to $100mL$ of a $0.01M$ solution of dichlorotetraaquachromium (III) chloride. The number of moles of $AgCl$ precipitated would be:

• A. $0.001$
• B. $0.002$
• C. $0.003$
• D. $0.01$

Answer 1

The correct option is A $0.001$

$\left[Cr\right(H_{2}O{)}_4\left(Cl{)}_2\right]Cl$ $\underrightarrow{excessAgN{O}_3}$ $\left[Cr\right(H_{2}O{)}_4(Cl{)}_2{]}^{+}+AgCl\downarrow$

Using formula, Molarity =$\frac{No.ofmoles}{Volume}\times 1000$

$0.01=$ $\frac{No.ofmoles}{100}\times 1000$

No. of moles of $AgCl=0.001$

Hence, the correct option is $A$