Question

A student wants to prepare 250 mL of a 0.25 mol/L lead(II) nitrate solution. Calculate the mass of lead(II) nitrate required for the preparation.

Answer 1

To prepare a 0.25 mol/L lead(II) nitrate solution with a volume of 250 mL, you will need 20.70 grams of lead(II) nitrate.

Step-by-step explanation:

To calculate the mass of lead(II) nitrate required, we need to use the formula:

moles = concentration (M) x volume (L)

Given:

Concentration = 0.25 mol/L

Volume = 250 mL = 0.250 L

First, we need to convert the volume from mL to L:

Volume = 0.250 L

Next, we can use the formula to calculate the moles of lead(II) nitrate:

moles = 0.25 mol/L * 0.250 L = 0.0625 moles

Finally, to calculate the mass of lead(II) nitrate, we can use the molar mass of Pb(NO3)2:

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

Mass of lead(II) nitrate = moles * molar mass

Mass = 0.0625 moles * 331.2 g/mol = 20.70 g