Question

A solution of antifreese is prepared by mixting 34.0mL of ethylene glycol (d=1.11g/mL; molar mass =62.07 g/mol) with 50.0 mL H20 (d=1.00 g/mol) at 25 degrees celsius. if the density of the antifreeze solution is 1.07 g/ mL, what is the molarity?

Answer 1

Answer: For an anntifreese Ethylene glycol,

Volume = 34.0 mL and Density = 1.11 g/mL.

Let's calculate mass of Ethylene Glycol using the relation,

Density = Mass / Volume

Mass = Density x Volume

Mass = 1.11 x 34 = 37.74 g.

Molar mass of Ethylene glycol = 62.07 g/mol

Let's calculate # of moles of ethylene glycol coresponds to mass 37.74 g.

# of moles = Given mass / Molar mass

# of moles of Ethylene glycol = 37.74 / 62.07 = 0.608 moles.

For a solvent water as density is 1 g/mL we have mass and volume same.

$\therefore$ Mass of water = 50 g and Volume of water = 50.0 mL

The antifreese solution is prepared by adding 34.0 mL i.e. 37.74 g ethylene glycol to 50.0 mL i.e. 50 g of water.

Mass of antifreese solution = 37.74 + 50.0 = 87.74 g

Expected volume of antifreese solution = 34.0 + 50.0 = 84.0 mL

But as ethylene glycol and water capable of showing strong H-bonding actual volume may be different.

Density of antifreese solution = 1.07 g/mL

From density of antifreese solution and mass of antifreese solution 87.74 g let's calculate actual volume of solution,

Actual volume of antifreese solution = Mass / Density = 87.74 / 1.07 = 82.0 mL = 0.082 L

Then,

Molarity = # of moles of ethylene glycol / Volume of solution in L = 0.608 / 0.082 = 7.41 M

Molarity of antifreese solution = 7.41 M.

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