7 grams of chloride was dissolved in water. 20 ml of lead(II)nitrate was used to completely precipitate all chloride (assume atomic mass of 35g/mol). What is the molarity of the Lead(II) Nitrate solution.

2Cl^- + Pb^2+ ==> PbCl2

mols Cl^- = grams/atomic mass Cl.
mols PbCl2 = 1/2 that.
mols Pb^2+ = mol PbCl2.
mols Pb(NO3)2 = mols PbCl2.
M Pb(NO3)2 = mols Pb(NO3)2/L soln.

DrBob222, Is the answer 5M?

Yes

thank you

To find the molarity of the Lead(II) Nitrate solution, we need to use the given information and some basic stoichiometry.

1. Start by calculating the number of moles of chloride present in the 7 grams using the formula:
Moles = Mass / Molar mass

Moles of chloride = 7 g / 35 g/mol = 0.2 mol

2. According to the reaction equation, 1 mole of lead(II) nitrate reacts with 2 moles of chloride:
Pb(NO3)2 + 2Cl- → PbCl2 + 2NO3-

Since the reaction is stoichiometric, we know that 0.2 mol of chloride will react with 0.1 mol of lead(II) nitrate.

3. Next, calculate the number of moles of lead(II) nitrate used:
Moles of lead(II) nitrate = 0.1 mol

4. Finally, calculate the molarity of the lead(II) nitrate solution:
Molarity = Moles of solute / Volume of solution (in liters)

Given volume of lead(II) nitrate = 20 ml = 0.02 L
Molarity = 0.1 mol / 0.02 L = 5 mol/L

Therefore, the molarity of the Lead(II) Nitrate solution is 5 mol/L.