Wednesday 13 February 2013

An antifreeze solution is prepared from 222.6 g of ethylene glycol (C2H6O2) and 200 g of water. Calculate the molality of the solution. If the density of the solution is 1.072 g mL–1, then what shall be the molarity of the solution?

Molecular mass of glycol C2H4(OH)2 =62
Number of moles of C2H4(OH)2 = mass of glycol / molar mass of glycol
= 222.6/62=3.59  moles
Molality of solution = number of moles of solute / mass of solvent in Kg  
Mass of solvent = 200g = 200/1000 Kg = 0.2 kg
Plug the values we get
Molality = 3.59/0.2
Molality = 17.95
Formula of molarity of solution = number of moles of solute / volume of solution in Kg
Formula of volume = mass / density
Volume = 422.6g/(1.072 g/ml)
Volume = 394.21 ml
Convert in liter   
Volume in liter = 394.21 ml /1000 liter  =0.394 liter
Molarity =9.1M

4 comments:

  1. Component present at largest quantity is solvent

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  2. Sir I have a confusion here amount of glycol is larger than water then glycol should be the solvent or not?

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