Explanation:
The volume of given lead nitrate solution is:
52.5 mL.
The amount of lead iodide formed is ---0.248 g.
To get the molarity of lead (II) ion follow the below-shown procedure:
The number of moles of lead iodide formed is:
[tex]number of moles of lead iodide(n)=mass of lead iodide/its molecular mass\\n=0.248 g/461.01g/mol\\n=0.000537mol[/tex]
0.000537 mole of lead iodide contains --- 0.000537 moles of lead (II) ion.
Thus, the number of moles are there, volume is there, and to get the molarity of lead (II) ion use the formula:
[tex]Molarity=\frac{number of moles}{volume in L.} \\M=0.000537 mol / 0.0525 L.\\M=0.0102mol/L[/tex]
Molarity of lead iodide is --- 0.0102 M.
