molarity = #mols/Liter of solution.
grams = #mols x molar mass.
So use equation 1, plug in M and L and calculate # mols. Then go to eqution 2 and plug in #mols and molar mass and calculate grams.
hi, i was wondering if anyone could help me with this problem? i am having some difficulty with it. thanks.
a).how many grams of Al(NO3)3 are needed to prepare 300mL of a 1.071 M solution?
b).if you further dilute this 300mL to a new final volume of 5.0L, calculate the final molarity.
3 answers
ok, i got this for a)68.376 g and b) 0.064 Molars. is it correct?
I calculated 68.43 g for (a) which round to 68.4 as does your answer. For b, I obtained 0.06426 which rounds to 0.0643 but the 5.0 L has only two significant figures; therefore, this would round to 0.064 M.
4 answers