Chemistry, asked by rohitpitti9046, 1 year ago

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?

Answers

Answered by santy2
3

Answer:

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?

Molality of the solution is 17.592 moles/kg

Molarity of the solution is 9.12 M

Explanation:

                     Molality  

Formula for calculating Molality is:

  Molality = Moles of solute / kilograms of solvent

Solute is Ethylene glycol while the solvent is water.

⇒Moles of Ethylene glycol:

             Moles = mass / molar mass

Mass of Ethylene glycol = 222.6 g

Molar mass of Ethylene glycol (C₂H₆O₂) = 24 + 6 + 32 = 62

Moles of Ethylene glycol (C₂H₆O₂)  = 222.6 / 62 = 3.59 moles

⇒ Mass of water as the solvent in kilograms = 200 g

200 g in kg = 200/1000 = 0.2 kg

∴ Molality   = 3.59 moles / 0.2kg

                 = 17.592 moles/kg

              Molarity  

Formula for molarity is:

                Molarity = Moles of solute / volume of solution in liters

Moles of the Ethylene glycol = 3.59 moles

Volume of the solution.

The denisty of the solution is 1.072 g/ml

The mass of the solution is 222.6 + 200 = 422.6g

Therefore the volume of the solution will be:

If 1.072 g = 1 ml

Then 422.6g = 1 ml × 422.6g/1.072g

                    = 394.23 ml

Convert 394.23 ml to liters using 1000ml = I liter conversion factor

This means that 394.23 ml = 394.23 ml/1000 Litres

                                          = 0.39423 litres

Molarity  = 3.59 moles / 0.39423 litres

              = 9.12 M

Answered by Anonymous
0

Explanation:

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

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