I'm finding this question a bit hard to understand. If someone could just explain what the question is asking for, and perhaps reword it, I would be happy to attempt it myself.
Methanol, CH3OH has a density of 0.787 g.cm^-3 at 25 degrees Celsius. A volumetric flask was found to be miscalibrated. The actual volume, when filled to the mark was 932.0 mL. If this flask is used to make a methanol solution of 0.800 molar, how much methanol (in ml) must be pipetted in, before it is diluted up to the mark?
okay so a density of 0.787 g.cm^3 means that for every cm^3, you have 0.787g of methanol.
Mr(CH3OH) = 32g/mol
so if we find out how many mols of methanol per cm^3 (aka mL, as 1cm^3 = 1mL),
n=m/Mr, n= 0.024594 mols
since the actual volume of the volumetric flask is 932.0 mL, a quick C=n/V can give us how many mols of methanol we need to pipette.
n= C x V, n = 0.8 x 0.932 n= 0.7456 mols of methanol required to make the flask have a 0.8M.
divide the amount of mols of methanol required by the amount of mols of methanol per mL.
0.7456/0.024594 = 30.3163 mL of methanol must be pipetted.
express to 3 sig figs, 30.3mL.