Answer :
The molarity of 10.1 grams of lead (IV) nitrate dissolved in 455.75 mL of water is 0.0669 M.
To calculate the molarity of 10.1 grams of lead (IV) nitrate dissolved in 455.75 mL of water, we will first convert grams to moles by using the molar mass of lead (IV) nitrate, Pb(NO3)2.
The molar mass of lead (IV) nitrate is 331.2 g/mol. Therefore, the number of moles of lead (IV) nitrate is:
Moles of Pb(NO3)2 = Mass (g) / Molar Mass (g/mol)
= 10.1 g / 331.2 g/mol
= 0.0305 mol.
Next, we convert the volume from milliliters to liters:
455.75 mL = 0.45575 L.
Now we can calculate the molarity (M) of the solution:
Molarity (M) = Moles of solute / Volume of solution (L)
= 0.0305 mol / 0.45575 L
= 0.0669 M.
The molarity of the lead (IV) nitrate solution is 0.0669 M.
