Question

Calculate the molality of ethanol solution in which the mole fraction of water is 0.88.

Answer 1

Mole fraction of water, X H₂O = 0.88

Mole fraction of ethanol, X C₂H₅OH = 1 - 0.88

= 0.12

X C₂H₅OH = n₂/n₁ + n₂    .......(1)

n₂ = number of moles of ethanol.

n₁ = number of moles of water. Molality of ethanol means the number of moles of ethanol present in 1000 g of water.

n₁ = 1000/18 = 55.5 moles

Substituting the value of n₁ in equation (1)

n₂/55.5 + n₂ = 0.12

n₂ = 7.57 moles 

Molality of ethanol ( C₂H₅OH) = 7.57 m

Alternatively,

Mole fraction of water = 0.88

Mole fraction of ethanol = 1-0.88 = 0.12

Therefore 0.12 moles of ethanol are present in 0.88 moles of water.

Mass of water = 0.88 x 18 =15.84 g of water.

Molality = number of moles of solute (ethanol) present in 1000 g of

solvent (water)

= 12 x 1000 / 15.84

= 7.57 m

Molality of ethanol ( C₂H₅OH) = 7.57 m