suppose 100 g of lead nitrate salt is added to enough water to make 1 L of solution, what is the concentration of the lead ion in the moles per liter in each of the solutions?
concentration (M) = moles of solute/liter of solution
convert 100g of lead nitrate salt to mole by dividing it by the molar mass (can be computed by looking at the periodic table)
divide mol of lead nitrate salt by 1 L of solution
To find the concentration of the lead ion in moles per liter (Molarity), we need to determine the number of moles of lead ions present in the solution and divide it by the volume of the solution in liters.
First, we need to calculate the number of moles of lead nitrate. We can use the molar mass of lead nitrate (Pb(NO3)2), which is the sum of the atomic masses of the elements in the compound.
Molar mass of Pb(NO3)2 = (207.2 g/mol for Pb) + (14.01 g/mol for N) + (3 * 16.00 g/mol for O)
Molar mass of Pb(NO3)2 = 207.2 g/mol + 14.01 g/mol + 48.00 g/mol
Molar mass of Pb(NO3)2 = 269.21 g/mol
Since we have 100 g of lead nitrate, we can calculate the moles of lead nitrate using the equation:
moles = mass / molar mass
moles of Pb(NO3)2 = 100 g / 269.21 g/mol
Next, we need to calculate the number of moles of lead ions (Pb2+) since each molecule of lead nitrate dissociates into one Pb2+ ion.
Since the ratio of lead ions to lead nitrate is 1:1, the moles of Pb(NO3)2 will be the same as the moles of Pb2+. Thus, the moles of Pb2+ = moles of Pb(NO3)2.
Now, we divide the moles of lead ions by the volume of the solution to get the concentration (Molarity) in moles per liter:
Concentration (Molarity) = moles of Pb2+ / volume of solution (in liters)
Since we have made 1 L of solution:
Concentration (Molarity) = moles of Pb2+ / 1 L
Therefore, the concentration of the lead ion in moles per liter would be equal to the moles of Pb2+ obtained from the previous calculation.