# Describe the steps, including the calculations, you would need to follow to prepare 150ml of a 0.25 mol/L solution of lead (II) nitrate. Be sure to provide enough detail in your process that a...

Describe the steps, including the calculations, you would need to follow to prepare 150ml of a 0.25 mol/L solution of lead (II) nitrate. Be sure to provide enough detail in your process that a student in this course could follow the steps to produce the solution.

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Molarity is defined as the amount of moles of solute per Liter of solution. It can also be expressed as:

`Molarity = (mol es of Solute)/(Volume (L))`

This problem is looking for the amount of lead nitrate to be used to prepare a solution of concentration at 0.25 `(mol)/(L)` . Rearranging the expression above we can have:

`mol es of Solute = (Molarity)*(Volume (L))`

- Molarity = 0.25 mol/L
- Volume = `(150/1000)L` = 0.15 L

`mol es of Solute = (0.25)*(0.15)`

`mol es of Solute` = 0.0375 moles lead (II) nitrate

To get the mass of the solute, multiply the moles acquired i the problem with the molecular mass of lead (II) nitrate.

`0.0375 mol es Pb(NO_3)_2 * (331.2 grams)/(1 mol ePb(NO_3)_2)`

`= 12.4 gramsPb(NO_3)_2`

To prepare the solution, measure 12.4 grams of `Pb(NO_3)_2` and place it in a flask. Add water until it reaches the 150 mL mark.