Question

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 19.40 g PbCl2(s) is obtained from 200.0 mL of the original solution.

Calculate the molarity of the Pb(NO3)2(aq) solution.

Answer 1

Step-by-step explanation:

To calculate the molarity of the Pb(NO3)2(aq) solution, you can use the information provided about the amount of PbCl2 precipitate formed. Here are the steps to calculate the molarity:

1. First, calculate the moles of PbCl2 precipitate formed. You know that 19.40 g of PbCl2 was obtained, so you can use its molar mass to find the moles:

Molar mass of PbCl2 = 207.2 g/mol (Pb) + 2 * 35.45 g/mol (Cl) = 277.1 g/mol

Moles of PbCl2 = 19.40 g / 277.1 g/mol ≈ 0.070 moles

2. Since the reaction between NaCl and Pb(NO3)2 is a 1:1 reaction (1 mole of PbCl2 forms from 1 mole of Pb(NO3)2), the moles of Pb(NO3)2 in the original solution are also approximately 0.070 moles.

3. Now, you need to find the volume of the original Pb(NO3)2(aq) solution. You were given that it was 200.0 mL.

4. Finally, calculate the molarity (M) using the formula:

Molarity (M) = Moles of solute / Volume of solution (in liters)

Molarity (Pb(NO3)2) = 0.070 moles / 0.200 L = 0.35 M

So, the molarity of the Pb(NO3)2(aq) solution is approximately 0.35 M.