Answer:
[tex][Cl^-]=232.3\frac{mgCl^-}{L}[/tex]
Explanation:
Hello,
In this case, we can represent the chemical reaction as:
[tex]Cl^-(aq)+AgNO_3(aq)\rightarrow AgCl(s)+NO_3^-(aq)[/tex]
In such a way, since the mass of the obtained silver chloride is 93.9 mg, we can compute the chloride ions in the ground water by using the following stoichiometric procedure whereas the molar mass of chloride ions and silver chloride are 35.45 g/mol and 143.32 g/mol respectively:
[tex]m_{Cl^-}=93.3mgAgCl*\frac{1mmolAgCl}{143.32mgAgCl}*\frac{1mmolCl^-}{1mmolAgCl} *\frac{35.45mgCl^-}{1mmolCl^-} =23.23mgCl^-[/tex]
Finally, for the given volume of water in liters (0.100L), we compute the required concentration:
[tex][Cl^-]=\frac{23.2mgCl^-}{0.100L}\\[/tex]
[tex][Cl^-]=232.3\frac{mgCl^-}{L}[/tex]
Best regards.
The concentration of chloride ions in the groundwater sample is 230 mg/L.
Based on the given information,
Now the number of moles of AgCl precipitate can be calculated as,
n = Given mass/Molar mass
Now putting the values we get,
[tex]n = \frac{0.939 g}{143.32 g/mol} \\n = 6.5 * 10^{-4}[/tex]
Thus, 6.5 × 10⁻⁴ moles of AgCl comprises 6.5 × 10⁻⁴ chloride ions. Therefore, 6.5 × 10⁻⁴ of chloride ions are present in the sample of 100 ml.
Now the molar mass of chloride ion is 35.453 g/mol, the mass of chloride ion will be,
Mass = Mole × Molar mass
Mass = 6.5 × 10⁻⁴ moles × 35.453 g/mol
Mass = 0.0230 g or 23 mg
The volume of the groundwater sample is 100 ml or 0.1000 L.
Now the concentration of the chloride ions in the sample given is,
C = 23 mg/0.1000 L
C = 230 mg/L
Thus, the concentration of chloride ions in the groundwater sample is 230 grams per liter.
Find out more information about the calculations of concentrations here:
https://brainly.com/question/6547081
