A solution of NaCl(aq) is added slowly to a solution of lead nitrate, Pb(NO3)2(aq) , until no further precipitation occurs. The precipitate is collected by filtration, dried, and weighed. A total of 18.86 g PbCl2(s) is obtained from 200.0 mL of the original solution. Calculate the molarity of the Pb(NO3)2(aq) solution.

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RomeliaThurston RomeliaThurston · Answer 1 · 2019-07-10T09:05:51+02:00

Answer: The molarity of [tex]Pb(NO_3)-2[/tex] solution is 0.34 M.

Explanation:

[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex]

For lead chloride:

Given mass of lead chloride = 18.86 g

Molar mass of lead chloride = 278.1 g/mol

Putting values in above equation, we get:

[tex]\text{Moles of lead chloride}=\frac{18.86g}{278.1g/mol}=0.068mol[/tex]

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

By Stoichiometry of the reaction:

1 mole of lead chloride is formed by 1 mole of lead nitrate

So, 0.068 moles of lead chloride will be formed from = [tex]\frac{1}{1}\times 0.068=0.068mol[/tex] of lead nitrate

[tex]\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}[/tex]

We are given:

Volume of solution = 200 mL = 0.200 L (Conversion factor: 1 L = 1000 mL)

Moles of lead nitrate = 0.068 moles

Putting values in above equation, we get:

[tex]\text{Molarity of }Pb(NO_3)_2\text{ solution}=\frac{0.068mol}{0.065L}\\\\\text{Molarity of }Pb(NO_3)_2\text{ solution}=0.34M[/tex]

Hence, the molarity of [tex]Pb(NO_3)-2[/tex] solution is 0.34 M.