Describe the steps, including 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 could follow the steps to produce the solution.
You are only allowed to ask one question at a time so I edited accordingly. We are looking to make 150 mL of a 0.25 M solution of lead nitrate. Basically, we need to figure out how many grams of lead nitrate are need to add to 150 mL of water to make a 0.25 M solution (M is molarity meaning moles per liter). Since we know the molarity of the final solution and the volume of final solution, we can multiply the two numbers to find the number of moles of lead nitrate needed. Remember that we need to convert 150 ml into liters (it is 0.150 L).
`(0.25 mol)/(L)*(0.150 L)=0.0375 mol`
We now need to convert this amount in moles of lead nitrate to grams of lead nitrate. We do this by multiplying by the molecular weight of lead nitrate (331.2 g/mol).
`0.0375 mol*(331.2 g)/(mol)=12.42 g`
So we now know that we need 12.42 grams of lead nitrate. So the procedure would be to weigh out 12.42 grams of lead nitrate on a balance and then add it to a beaker. Then add water to a graduated cylinder until it reaches the 150 mL mark. Pour the water from the graduated cylinder into the beaker with the lead nitrate. Stir it until fully dissolved.