You are supposed to mix 250 mL of a 0.1 M solution of Pb(NO3)2 solution. You would need to mix ______ g of Pb(NO3)2 with enough water to make 250 mL of solution.
How many mols do you need? That's M x L = ?
Then grams = mols x molar mass
Thank you for the help
To determine the mass of Pb(NO3)2 required to make a 0.1 M solution in 250 mL, we can use the formula:
Molarity (M) = moles of solute (mol) / volume of solution (L)
First, let's find the number of moles of Pb(NO3)2 required:
Given:
Molarity = 0.1 M
Volume of solution = 250 mL = 0.25 L
Using the formula, we can rearrange it to solve for moles of solute:
moles of solute = Molarity × volume of solution
moles of solute = 0.1 M × 0.25 L
moles of solute = 0.025 mol
Now, to find the mass of Pb(NO3)2 required, we need to multiply the moles by the molar mass of Pb(NO3)2, which can be obtained from the periodic table:
Pb(NO3)2:
Atomic mass of Pb = 207.2 g/mol
Atomic mass of N = 14.0 g/mol
Atomic mass of O = 16.0 g/mol (x3)
Molar mass of Pb(NO3)2 = (207.2 g/mol) + (14.0 g/mol) + (16.0 g/mol × 3) = 331.2 g/mol
Finally, we can find the mass of Pb(NO3)2:
mass = moles of solute × molar mass
mass = 0.025 mol × 331.2 g/mol
mass = 8.28 g (rounded to two decimal places)
Therefore, you would need to mix approximately 8.28 grams of Pb(NO3)2 with enough water to make 250 mL of solution to obtain a 0.1 M concentration.