Final answer:
To obtain 13.6 grams of lead acetate, you would need approximately 363 milliliters of a 0.181 M aqueous solution of lead acetate.
Step-by-step explanation:
To find the volume of an aqueous solution of lead acetate needed to obtain 13.6 grams of the salt, we need to use the formula:
Molarity (M) = Moles (mol) / Volume (L)
First, we need to convert the given mass of lead acetate to moles using its molar mass.
Molar Mass of Pb(CH3COO)2 = 207.2 g/mol
Moles = Mass / Molar Mass = 13.6 g / 207.2 g/mol = 0.0657 mol
Next, we can use the formula rearranged to solve for volume:
Volume (L) = Moles (mol) / Molarity (M)
According to the information provided, the molarity of the lead acetate solution is 0.181 M.
Volume (L) = 0.0657 mol / 0.181 M = 0.363 L
Finally, we need to convert the volume from liters to milliliters:
Volume (mL) = Volume (L) * 1000 mL/L = 0.363 L * 1000 = 363 mL