posted by T on .
it is desired to prepare 250.0mL of a standard solution having a concentration of 0.1000M AgNo3. How many grams of a sample of AgNO3 of 98.8% purity are required for this purpose?
M = moles/L
0.1 = moles/0.250 L
solve for moles.
moles = grams/molar mass
You know moles and molar mass, solve for grams (100% pure) AgNO3.
grams impure AgNO3 x 0.988 = grams pure AgNO3 you want or rearrange to
grams pure AgNO3/0.988 = grams you must weigh out to get grams AgNO3 you need.
the answer in the book was 4.3 grams
but if you do that you get
169.8731/0.988 which is 171.94 g
i don't get it