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A solution of NaCl ( aq ) is added slowly to a solution of lead nitrate, Pb ( NO 3 ) 2 ( aq ) , until no further precipitation occurs. The precipitate is collected by filtration, dried, and weighed. A total of 19.58 g CaCl2 ( s ) is obtained from 200.0 mL of the original solution. Calculate the molarity of the Pb ( NO 3 ) 2 ( aq ) solution.

Answer :

Answer : The molarity of [tex]Pb(NO_3)_2[/tex]  solution is, 0.352 M

Explanation :

First we have to calculate the moles of [tex]PbCl_2[/tex]

[tex]\text{Moles of }PbCl_2=\frac{\text{Given mass }PbCl_2}{\text{Molar mass }PbCl_2}[/tex]

Molar mass of [tex]PbCl_2[/tex] = 278.1 g/mol

[tex]\text{Moles of }PbCl_2=\frac{19.58g}{278.1g/mol}=0.07041mol[/tex]

Now we have to calculate the moles of [tex]CaCl_2[/tex]

The balanced chemical equation is:

[tex]Pb(NO_3)_2(aq)+2NaCl(aq)\rightarrow PbCl_2(s)+2NaNO_3(aq)[/tex]

From the balanced reaction we conclude that

As, 1 mole of [tex]PbCl_2[/tex] produced from 1 mole of [tex]Pb(NO_3)_2[/tex]

So, 0.07041 mole of [tex]PbCl_2[/tex] produced from 0.07041 mole of [tex]Pb(NO_3)_2[/tex]

Now we have to calculate the molarity of [tex]Pb(NO_3)_2[/tex]

[tex]\text{Molarity of }Pb(NO_3)_2=\frac{\text{Moles of }Pb(NO_3)_2}{\text{Volume of solution in (L)}}[/tex]

[tex]\text{Molarity of }Pb(NO_3)_2=\frac{0.07041mol}{0.200L}=0.352M[/tex]

Therefore, the molarity of [tex]Pb(NO_3)_2[/tex]  solution is, 0.352 M

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