35 % (W/W) solution of ethylene glycol in water, an anti-freezer used

1.	35 % (W/W) solution of ethylene glycol in water, an anti-freezer used in automobiles in radiators as a coolant. It lowers freezing point of water to -17.6 °C. Calculate the mole fraction of the components.
Answer» 35% (W/W) means 100 g solution contains 35 g ethylene glycol (CH₂OH-CH₂OH) and 65 g H₂O. Molar mass of water = 18 g mol^-1 Molar mass of ethylene glycol (CH₂OHCH₂OH) = 62 g mol^-1 Number of moles of water = n₁ = \(\frac{65}{18}\)= 3.611 mol Number of moles of ethylene glycol = n₂ = \(\frac{35}{62}\)= 0.5645 mol Total moles = n = n₁ + n₂ = 3.611 + 0.5645 = 4.1755 mol Mole fraction of ethylene glycol = x₂ = \(\frac{n_2}{n_1}\)= \(\frac{0.5645}{4.1755}\) = 0.1352 ∴ Mole fraction of water = 1 – x₂ = 1 – 0.1352 = 0.8648 ∴ Mole fraction of water = 0.8648 Mole fraction of ethylene glycol = 0.1352

35 % (W/W) solution of ethylene glycol in water, an anti-freezer used in automobiles in radiators as a coolant. It lowers freezing point of water to -17.6 °C. Calculate the mole fraction of the components.

Answer»

35% (W/W) means 100 g solution contains 35 g ethylene glycol (CH₂OH-CH₂OH) and 65 g H₂O.

Molar mass of water = 18 g mol^-1

Molar mass of ethylene glycol (CH₂OHCH₂OH) = 62 g mol^-1

Number of moles of water = n₁ = \(\frac{65}{18}\)= 3.611 mol

Number of moles of ethylene glycol = n₂ = \(\frac{35}{62}\)= 0.5645 mol

Total moles = n = n₁ + n₂

= 3.611 + 0.5645

= 4.1755 mol

Mole fraction of ethylene glycol = x₂

= \(\frac{n_2}{n_1}\)= \(\frac{0.5645}{4.1755}\) = 0.1352

∴ Mole fraction of water = 1 – x₂ = 1 – 0.1352

= 0.8648

∴ Mole fraction of water = 0.8648 Mole fraction of ethylene glycol = 0.1352

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