Calculate the molality of ethanol solution in which the mole fraction

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

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

Answer»

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

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