Hello again, I need some help ONCE AGAIN...
The quesiton is as follows..
In a lab experiment when 20ml of Lead (II) Nitrate is mixed with 40ml of Potassium Iodide and then strained through a buncher funnel there is 9.112 grams of Lead(II) Iodide precipitate, with 0.108g of Lead(II) Iodide left over along with 4.043796g of Potassium Nitrate and 60g of Water.
The question is
A) what is the molar mass of Lead(II) Iodide ( I got 253.8g/mol)
B)What is the theoretical yield of lead (II) Iodide using the following equation...
Theo yield = 20ml X 1L/1000mL X 1.00 Mol/L X molar mass of PbI2
What I have so far is
Theo Yield = 20ml PbI2 X 1.00 mol PbI2/253.8ml PbI2
= 0.08mol/mL PbI2
Now where I am unsure is the point where 1.00mol/L. If mol are measured in grams than how can a liquid unit of a mol be measure in Litres? Anyways its probably something silly.
C)calculate the percentage yield of solid lead(II) iodide.
I know that the percentage yield = actual yield/thoretical yield but because I am unsure as to whether or not my theoretical yield is correct i cannot be sure the percentage yield is correct
Please help me in anyway you can, this site has been a great help in the past and I hope to hear form somebody soon.