[Danielle]

1. How many grams of solute is required to prepare 1.000 L of 1.000 M Pb(NO3)2? What is the molar concentration of the solution for each of the ions?

 

[UG]

Can you please show an attempt so that we know what you are having trouble with?

[mehc]

1.00 M lead (II) nitrate solution mean one mole of Pb(NO₃)₂ in one litre of water (Solvent). One mole of Pb(NO₃)₂ is equal to 331.2 g.

In order to calculate individual ions molarity, use coefficients
from the balanced dissolution equation.
Pb(NO₃)₂  = Pb²⁺  + 2NO₃^-

e.g. for nitrate ions molarity
(1 mole Pb(NO₃)₂ /1 liter soln) x(2 mole NO₃^-/1 mole Pb(NO₃)₂ )
= 2 mole NO₃^- / 1 L soln. or 2 M NO₃^-

Hope it will help.