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3 years ago

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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.

1 answer:

Hoochie [10]3 years ago

7 0

<u>Answer:</u> The molarity of $Pb(NO_3)-2$ solution is 0.34 M.

<u>Explanation:</u>

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$

<u>For lead chloride:</u>

Given mass of lead chloride = 18.86 g

Molar mass of lead chloride = 278.1 g/mol

Putting values in above equation, we get:

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

For the balanced chemical equation:

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

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 = $\frac{1}{1}\times 0.068=0.068mol$ of lead nitrate

To calculate the molarity of solution, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}$

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:

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

Hence, the molarity of $Pb(NO_3)-2$ solution is 0.34 M.

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