aqueous solution of ammonium sulfate.Calculate the final molarity of lead(II) cation in the solution. You can assume the volume of the solution doesn't change when the lead(II) nitrate is dissolved in it.Be sure your answer has the correct number of significant digits.

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the no. of moles of lead in the solution is :

[25.4 /331]

= 0.076 moles

the final volume of the solution is : 300 ml = 0.3 lit

so the molarity will be 0.076/0.3

=0.25M

so the molarity of lead (II) ions is 0.25M Please let me know if you need any clarification for the doubts. I'm always happy to answer your questions. Looking forward to help you again. thank you. :)