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luda_lava [24]
3 years ago
13

Calculate the mass of ethylene glycol (C2H6O2) that must be added to 1.00 kg of ethanol (C2H5OH) to reduce its vapor pressure by

12.0 torr at 35∘C. The vapor pressure of pure ethanol at 35∘C is 1.00×102 torr.
Chemistry
1 answer:
Diano4ka-milaya [45]3 years ago
4 0

Answer:

183.7 g of ethylene glycol

Explanation:

This a case of colligative property, about vapour pressure lowering.

ΔPv = P° . Xst

ΔPv = 12 Torr

P° = Vapour pressure of pure solvent → 100 Torr

12  Torr = 100 Torr . Xst

12 Torr / 100 Torr = 0.12 → Xst

Xst is mole fraction for solute which means moles of solute / Total moles and, the total moles are the sum of moles of solvent + moles of solute.

We can make this equation:

Moles of solute / (Moles of solute + Moles of solvent) = 0.12

We don't know the moles of solute, but we do know the moles of solvent by the mass.

Mass / Molar mass = Mol

1000 g / 46 g/m = 21.7 moles

Moles of solute / Moles of solute + 21.7 moles = 0.12

Moles of solute = 0.12 ( Moles of solute + 21.7 moles)

Moles of solute - 0.12 moles of solute = 2.60  moles

0.88 moles of solute = 2.60 moles

Moles of solute = 2.60 moles / 0.88 → 2.96 moles of solute

Mol . molar mass = Mass

2.96 m  .  62g/m = 183.7 g

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The splitting of d orbitals according to the crystal field theory depends on the;

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When the geometry and the ligands are held constant, the order of crystal field splitting is as follows;

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[Ru(CN)6]3- will exhibit a greater crystal field splitting. Crystal field splitting increases with the second and third row transition elements when compared to the crystal field splitting of the first row transition elements. Note that, there is an increase of approximately 30%–50% in Δo on going from a first-row transition metal to a second-row metal and another 30%–50% increase on going from a second-row to a third-row metal when they have the same geometry and oxidation state.

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