Answer:
[tex]M=0.0637M[/tex]
Explanation:
Hello,
In this case, the undergoing chemical reaction is:
[tex]Pb(NO_3)_2(aq)+2NaI(aq)\rightarrow PbI_2(s)+2NaNO_3(aq)[/tex]
Thus, for 0.904 g of precipitate, that is lead (II) iodide, we can compute the initial moles of lead (II) ions in lead (II) nitrate:
[tex]n_{Pb^{2+}}=0.904gPbI_2*\frac{1molPbI_2}{461gPbI_2}*\frac{1molPb(NO_3)_2}{1molPbI_2} *\frac{1molPb^{2+}}{1molPb(NO_3)_2} =1.96x10^{-3}molPb^{2+}[/tex]
Finally, the resulting molarity in 30.8 mL (0.0308 L):
[tex]M=\frac{1.96x10^{-3}molPb^{2+}}{0.0308L}\\ \\M=0.0637M[/tex]
Regards.
