To calculate the concentration of tin(II) chloride contaminant in the original groundwater sample, we can follow these steps:
1. Convert the mass of silver chloride from mg to grams:
[tex]\[
3.8 \text{ mg} = \frac{3.8}{1000} \text{ g} = 0.0038 \text{ g}
\][/tex]
2. Convert the volume of the groundwater sample from mL to L:
[tex]\[
250 \text{ mL} = \frac{250}{1000} \text{ L} = 0.25 \text{ L}
\][/tex]
3. Calculate the molar mass of silver chloride (AgCl):
- Molar mass of Ag: [tex]\(107.8682 \, \text{g/mol}\)[/tex]
- Molar mass of Cl: [tex]\(35.45 \, \text{g/mol}\)[/tex]
[tex]\[
\text{Molar mass of AgCl} = 107.8682 \, \text{g/mol} + 35.45 \, \text{g/mol} = 143.3182 \, \text{g/mol}
\][/tex]
4. Calculate the moles of silver chloride (AgCl):
[tex]\[
\text{Moles of AgCl} = \frac{0.0038 \, \text{g}}{143.3182 \, \text{g/mol}} \approx 2.6514427337211885 \times 10^{-5} \, \text{moles}
\][/tex]
5. Determine the moles of tin(II) chloride (SnCl[tex]\(_2\)[/tex]) using the balanced chemical equation:
- From the balanced equation, 1 mole of SnCl[tex]\(_2\)[/tex] reacts with 2 moles of AgNO[tex]\(_3\)[/tex], producing 2 moles of AgCl.
- Therefore, the moles of SnCl[tex]\(_2\)[/tex] is half the moles of AgCl:
[tex]\[
\text{Moles of SnCl}_2 = \frac{2.6514427337211885 \times 10^{-5}}{2} \approx 1.3257213668605943 \times 10^{-5} \, \text{moles}
\][/tex]
6. Calculate the concentration of tin(II) chloride (SnCl[tex]\(_2\)[/tex]) in the groundwater sample:
[tex]\[
\text{Concentration of SnCl}_2 = \frac{\text{Moles of SnCl}_2}{\text{Volume of groundwater sample in L}} = \frac{1.3257213668605943 \times 10^{-5}}{0.25} \approx 5.302885467442377 \times 10^{-5} \, \text{M}
\][/tex]
7. Convert this concentration to millimolar (mM):
[tex]\[
\text{Concentration of SnCl}_2 \text{ in mM} = 5.302885467442377 \times 10^{-5} \times 1000 \approx 0.053028854674423774 \, \text{mM}
\][/tex]
8. Round the answer to 2 significant digits:
[tex]\[
\text{Concentration of SnCl}_2 = 0.053 \, \text{mM}
\][/tex]
Therefore, the concentration of tin(II) chloride contaminant in the original groundwater sample is approximately [tex]\(0.053\)[/tex] mM.
