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.
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