# how to prepare 150ml of a 0.25 mol/L solution of lead (II) nitrate. Be sure to provide enough detail in your process to produce the solutionDescribe the steps, including the calculations

jerichorayel | College Teacher | (Level 2) Senior Educator

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To prepare a solution of Lead (II) nitrate, first we have to deal with the calculations.

Since the concentration is given, we can start from that point.

Concentration = moles of solute/Volume of the solution (L)

Concentration = 0.25 mol/L

Volume = (150/1000)L = .15L

By arranging the formula we can have:

moles of solute = Concentration * Volume

moles of solute = 0.25 * .15

moles of solute = 0.0375 moles Lead (II) nitrate

Next we convert the moles of Lead (II) nitrate (Pb(NO3)2) to mass. We can do that by multiplying the moles with the molar mass of Pb(NO3)2

mass Pb(NO3)2 = 0.0375mol * 331.2 g/mol

mass Pb(NO3)2 = 12.42 grams Pb(NO3)2

In preparing the solution, first, measure or weigh 12.42 grams of Pb(NO3)2. Transfer the measured mass in a flask and add water until it reaches the 150mL mark.