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77julia77 [94]
3 years ago
6

How many grams of antifreeze C2H4(OH)2 would be required per 500 g of water to prevent the water from freezing at a temperature

of -20.0 C
Chemistry
1 answer:
Wewaii [24]3 years ago
5 0

Answer:

333.7 g.

Explanation:

  • The depression in freezing point of water (ΔTf) due to adding a solute to it is given by: <em>ΔTf = Kf.m.</em>

Where, ΔTf is the depression in water freezing point (ΔTf = 20.0°C).

Kf is the molal freezing point depression constant of the solvent (Kf = 1.86 °C/m).

m is the molality of the solution.

<em>∴ m = ΔTf/Kf</em> = (20.0°C)/(1.86 °C/m) = <em>10.75 m.</em>

molaity (m) is the no. of moles of solute per kg of the solvent.

∵ m = (no. of moles of antifreeze C₂H₄(OH)₂)/(mass of water (kg))

∴ no. of moles of antifreeze C₂H₄(OH)₂ = (m)(mass of water (kg)) = (10.75 m)(0.5 kg) = 5.376 mol.

∵ no. of moles = mass/molar mass.

<em>∴ mass of antifreeze C₂H₄(OH)₂ = no. of moles x molar mass </em>= (5.376 mol)(62.07 g/mol) =<em> 333.7 g.</em>

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