Question

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 1

Answer : The molarity of $Pb(NO_3)_2$  solution is, 0.352 M

Explanation :

First we have to calculate the moles of $PbCl_2$

$\text{Moles of }PbCl_2=\frac{\text{Given mass }PbCl_2}{\text{Molar mass }PbCl_2}$

Molar mass of $PbCl_2$ = 278.1 g/mol

$\text{Moles of }PbCl_2=\frac{19.58g}{278.1g/mol}=0.07041mol$

Now we have to calculate the moles of $CaCl_2$

The balanced chemical equation is:

$Pb(NO_3)_2(aq)+2NaCl(aq)\rightarrow PbCl_2(s)+2NaNO_3(aq)$

From the balanced reaction we conclude that

As, 1 mole of $PbCl_2$ produced from 1 mole of $Pb(NO_3)_2$

So, 0.07041 mole of $PbCl_2$ produced from 0.07041 mole of $Pb(NO_3)_2$

Now we have to calculate the molarity of $Pb(NO_3)_2$

$\text{Molarity of }Pb(NO_3)_2=\frac{\text{Moles of }Pb(NO_3)_2}{\text{Volume of solution in (L)}}$

$\text{Molarity of }Pb(NO_3)_2=\frac{0.07041mol}{0.200L}=0.352M$

Therefore, the molarity of $Pb(NO_3)_2$  solution is, 0.352 M