100-31.0=69.0g, or better yet, 0.069 kg of solvent.
31g of C2H6O2*(1 mole/62.06810 g)= moles of C2H6O2
m=molality=moles of C2H6O2/0.069 kg of solvent
An aqueous antifreeze solution is 31.0% ethylene glycol (C2H6O2) by mass.
The density of the solution is 1.039 g/cm3.
Calculate the molality of the ethylene glycol.
3 answers
31% means 31 g glycol/100 g soln.
Convert 31 g glycol to mols. mol = grams/molar mass
Convert 100 g solution to g solute + g solvent. Convert g H2O to kg, then m = mols/kg solvent.
Convert 31 g glycol to mols. mol = grams/molar mass
Convert 100 g solution to g solute + g solvent. Convert g H2O to kg, then m = mols/kg solvent.
Awesome! Thank you
1 answer