You made a 100.0 ml of a lead(ii) nitrate solution for lab but forgot to cap it. the next lab session you noticed that there was only 80.0 ml left (the rest had evaporated). in addition, you forgot the initial concentration of the solution. you decide to take 2.00 ml of the solution and add an excess of a concentration sodium chloride solution. you obtain a solid with a mass of 3.407 g. what was the concentration of the original lead(ii) nitrate solution?

Answer is:  concentration of the original lead(II) nitrate solution is 4,88 mol/l.
Chemical reaction: Pb(NO₃)₂ + 2NaCl → PbCl₂ + 2NaNO₃.
V(Pb(NO₃)₂) = 2 ml = 0,002 l.
m(PbCl₂) = 3,407 g.
n(PbCl₂) = m(PbCl₂) ÷ M(PbCl₂).
n(PbCl₂) = 3,407 g ÷ 278,1 g/mol.
n(PbCl₂) = 0,0122 mol.
From chemical reaction: n(PbCl₂) : n(Pb(NO₃)₂) = 1 : 1.
n(Pb(NO₃)₂) = 0,0122 mol.
c(Pb(NO₃)₂) = n(Pb(NO₃)₂) ÷ V(Pb(NO₃)₂).
c(Pb(NO₃)₂) = 0,0122 mol ÷ 0,002 l.
c(Pb(NO₃)₂) = 6,1 mol/l in 80 ml.
c(Pb(NO₃)₂) = 6,1 mol/l · 0,8 = 4,88 mol/l.

Wish i can but don’t know this

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