Answer:
The original solution was 0.20 M AgNO₃.
Explanation:
We can first write the balanced chemical equation of the double-replacement reaction between the NaCl solution and AgNO₃ solution:
[tex]\displaystyle \text{NaCl$_\text{(aq)}$ $+$ AgNO$_{3\text{(aq)}}$} \longrightarrow \text{NaNO$_{3\text{(aq)}}$ $+$ AgCl$_\text{(aq)}$}[/tex]
Recall that molarity is given by:
[tex]\displaystyle \text{M} = \frac{\text{ mols solute}}{\text{ L soln.}}[/tex]
To find the molarity of the original AgNO₃ solution, then, we will need to find the amount of AgNO₃ in moles. We are given that the volume of the original solution was 38 mL.
To determine the moles of AgNO₃, we can convert the mass of AgCl to moles of AgCl and then to moles of AgNO₃.
We are given that the formula weight of AgCl is 143.32 g/mol.
From the equation, we can see that one mole of AgNO₃ yields one mole of AgCl.
This yields two ratios:
[tex]\displaystyle \frac{1\text{ mol AgCl}}{143.32 \text{ g AgCl}} \text{ and } \frac{1\text{ mol AgNO$_3$}}{1 \text{ mol AgCl}}[/tex]
Using the above ratios, multiply with the given mass of AgCl:
[tex]\displaystyle 1.11 \text{ g AgCl} \cdot \frac{1\text{ mol AgCl}}{143.32\text{ g AgCl}} \cdot \frac{1 \text{ mol AgNO$_3$}}{1\text{ mol AgCl}} = 0.00774\text{ mol AgNO$_3$}[/tex]
Therefore, the molarity of the original AgNO₃ solution is:
[tex]\displaystyle \text{M} & = \frac{(0.00774 \text{ mol AgNO$_3$})}{(38\text{ mL soln.})}\cdot \frac{1000 \text{ mL soln.}}{1\text{ L soln.}} = 0.20 \text{ M}[/tex]
In conclusion, the original solution was 0.20 M AgNO₃.
