Respuesta :
Answer: The concentration of chloride ions in the solution is 1.056 mol/L
Explanation:
The chemical equation for the reaction of lead (II) nitrate and magnesium chloride follows:
[tex]Pb(NO_3)_2+MgCl_2\rightarrow PbCl_2+Mg(NO_3)_2[/tex]
All the chloride ions are getting converted to lead (II) chloride
To calculate the number of moles, we use the equation:
[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex]
Given mass of lead (II) chloride = 7.35 g
Molar mass of lead (II) chloride = 278.1 g/mol
Putting values in above equation, we get:
[tex]\text{Moles of }PbCl_2=\frac{7.35g}{278.1g/mol}=0.0264mol[/tex]
1 mole of lead (II) chloride produces 1 mole of lead ions and 2 moles of chloride ions.
So, moles of chloride ions = (2 × 0.0264) = 0.0528 moles
To calculate the molarity of solution, we use the equation:
[tex]\text{Molarity of chloride ions in solution}=\frac{\text{Moles of chloride ions}\times 1000}{\text{Volume of solution (in mL)}}[/tex]
Moles of chloride ions = 0.0528 moles
Volume of solution = 50.0 mL
Putting values in above equation, we get:
[tex]\text{Concentration of chloride ions in solution}=\frac{0.0528\times 1000}{50.0}\\\\\text{Concentration of chloride ions in solution}=1.056mol/L[/tex]
Hence, the concentration of chloride ions in the solution is 1.056 mol/L
