How many grams of ethylene glycol, HOCH2CH2OH, are present in 5.3 gallons of a 5.22 m solution with a density of 1.83 g/mL?

Answer 1

I would do this.
5.22m means 5.22 moles in 1 kg solvent. If we take 1 kg solvent we will have 5.22 x molar mass ethylene glycol = 5.22*62 = about 324 g Ethgly. The solution will have a mass of 1000 g + 324 = about 1324 (You need to go through and redo since I've estimated here and there.). So how many grams ethylene glycol must we have in a 5.3 gallon. I typed 5.3 gallons to liters into google and it returned 20.063 L or 20063 mL and that x 1.83 gives a mass of about 36,716 grams for the solution.
So moles ethylene glycol = 324 x (36715/1324) = = about 8985 g ethylene glycol.
I like to check these things to see if it will make a 5.22 m soln.
8984/62 = 145 moles ethylene glycol.
mass soln = 36715 and subtract 8984 g for the solute = about 27,700 g or 27.7 kg. Then 145 moles/27.7 kg = 5.23m which is a good answer considering I rounded here and there.