Suppose we have a solution of lead nitrate, Pb(NO3). A solution of NaCl(aq) is added slowly until no further precipitation occurs. The precipitate is collected by filtration, dried and weighed. A total of 10.27g of PbCl2(s) is obtained from 200.0mL of the original solution.
Calculate the molarity of Pb(NO3)2(aq)solution.
To calculate the molarity of the Pb(NO3)2(aq) solution, we need to know the number of moles of Pb(NO3)2 and the volume of the solution.
1. Calculate the number of moles of PbCl2:
We know that the molar mass of PbCl2 is 278.1 g/mol (207.2 g/mol for Pb plus 2 * 35.5 g/mol for Cl).
So, the number of moles of PbCl2 can be calculated using the formula:
Moles = Mass / Molar mass
Moles of PbCl2 = 10.27 g / 278.1 g/mol
2. Calculate the number of moles of Pb(NO3)2:
Since lead nitrate (Pb(NO3)2) and lead chloride (PbCl2) have a 1:2 stoichiometric ratio, the number of moles of Pb(NO3)2 will be twice the number of moles of PbCl2.
Moles of Pb(NO3)2 = 2 * Moles of PbCl2
3. Calculate the volume of Pb(NO3)2 solution:
We are given that the original solution was 200.0 mL. Therefore, the volume of the Pb(NO3)2 solution is 200.0 mL.
4. Calculate the molarity (M) of Pb(NO3)2 solution:
Molarity = Moles / Volume (in liters)
Divide the number of moles of Pb(NO3)2 by the volume of the solution in liters:
Molarity = Moles of Pb(NO3)2 / Volume (in liters)
Convert the volume from milliliters to liters:
Volume (in liters) = 200.0 mL / 1000
Substitute the values in the formula:
Molarity = (2 * Moles of PbCl2) / (200.0 mL / 1000)
Calculate the molarity using the given values.
By following these steps, you can calculate the molarity of the Pb(NO3)2(aq) solution.
10.27g PbCl2/ molar mass PbCl2 = ? mols PbCl2.
That means that many mols Pb(NO3)2. That was in 200 mL solution;M = mols/L.