I typed in 2.6 gallons to liters in google and it returned 9.842 liters and that is 9842 mL. That will have a mass of 9842 mL x 1.88 g/mL = 18503 grams
If we take 1 kg of solution that will contain 4.61 moles ethylene glycol and that is 4.61 x 62.07 g/mol = about 286 grams which makes the solution have a mass of 1000 g + 286 g = 1286 grams. (I'm rounding here and there and estimating so you need to redo all of the calculations and keep in mind the number of significant figures.)
So how many grams ethylene glycol will we need in the 18503 g soln? That will be
286 g x (18503/1286) = about 4100 g.
I like to check these things to see if 4100 g will give 4.61 m in 18503g soln. If may not be exact because I've rounded here and there.
moles solute = 4100/62 = 66 moles.
Grams soln = 18503 - 4100 g solute = 14,400 g solvent which is 14.4 kg solvent. Then 66 moles/14.4 kg solvent = 4.58 which is very close considering the approximations made.
How many grams of ethylene glycol, HOCH2CH2OH, are present in 2.6 gallons of a 4.61 m solution with a density of 1.88 g/mL?
i think that this is a dimensional analysis type of question. 1 gal ~ 3485.41 ml.
i just need help with the set up, i think...
1 answer