Question

Calculate the mass of lead(II) nitrate required to prepare a 250 mL, 0.25 mol/L solution, respecting the user's request not to provide solutions.

Answer 1

To prepare a 250 mL, 0.25 mol/L solution of lead (II) nitrate, you would need to use 20.7 grams of the compound.

Step-by-step explanation:

To calculate the mass of lead(II) nitrate required to prepare a 250 mL, 0.25 mol/L solution, we can use the formula:

mass = molarity x volume x molar mass

Molar mass of Pb(NO3)2 = 331.2 g/mol

Plugging in the values, mass = 0.25 mol/L x 0.250 L x 331.2 g/mol = 20.7 g

Therefore, you would need to use 20.7 grams of lead(II) nitrate to prepare the given solution.