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Calculate the mass of a non-volatile solute (molar mass 40 g mol-1) which should be dissolved in 114 g octane to reduce its vapour pressure to 80%.
- a)8 g
- b)5 g
- c)106.2 g
- d)None of these.
Correct answer is option 'A'. Can you explain this answer?
Verified Answer
Calculate the mass of a non-volatile solute (molar mass 40 g mol-1) wh...
Let the vapour pressure of pure octane be p10.
Then, the vapour pressure of the octane after dissolving the non-volatile solute is 80/100 p10 = 0.8 p10.
Molar mass of solute, M2 = 40 g mol - 1
Mass of octane, w1 = 114 g
Molar mass of octane, (C8H18), M1 = 8 × 12 + 18 × 1 = 114 g mol - 1
Applying the relation,
(p10 - p1) / p10 = (w2 x M1 ) / (M2 x w1 )
⇒ (p10 - 0.8 p10) / p10 = (w2 x 114 ) / (40 x 114 )
⇒ 0.2 p10 / p10 = w2 / 40
⇒ 0.2 = w2 / 40
⇒ w2 = 8 g
Hence, the required mass of the solute is 8 g.
Most Upvoted Answer
Calculate the mass of a non-volatile solute (molar mass 40 g mol-1) wh...
Given:
Molar mass of non-volatile solute = 40 g/mol
Mass of octane = 114 g
Vapour pressure reduction = 80%
To find: Mass of non-volatile solute required
Formula used: Raoult's law
According to Raoult's law, the vapour pressure of a solution is directly proportional to the mole fraction of the solvent present in the solution.
P(total) = P(solvent) x mole fraction of solvent
Let the mole fraction of the solvent be x and the mole fraction of the solute be (1-x).
P(total) = P(octane) x x (as octane is the solvent)
0.8P(octane) = P(octane) x x
x = 0.8
Therefore, the mole fraction of the solute = (1 - 0.8) = 0.2
Mole fraction of solute = mass of solute / (molar mass of solute x total mass of solution)
Mass of solute = Mole fraction of solute x molar mass of solute x total mass of solution
Mass of solute = 0.2 x 40 g/mol x 114 g
Mass of solute = 8 g
Therefore, the mass of the non-volatile solute required to reduce the vapour pressure of 114 g of octane to 80% is 8 g. Option (a) is the correct answer.
Molar mass of non-volatile solute = 40 g/mol
Mass of octane = 114 g
Vapour pressure reduction = 80%
To find: Mass of non-volatile solute required
Formula used: Raoult's law
According to Raoult's law, the vapour pressure of a solution is directly proportional to the mole fraction of the solvent present in the solution.
P(total) = P(solvent) x mole fraction of solvent
Let the mole fraction of the solvent be x and the mole fraction of the solute be (1-x).
P(total) = P(octane) x x (as octane is the solvent)
0.8P(octane) = P(octane) x x
x = 0.8
Therefore, the mole fraction of the solute = (1 - 0.8) = 0.2
Mole fraction of solute = mass of solute / (molar mass of solute x total mass of solution)
Mass of solute = Mole fraction of solute x molar mass of solute x total mass of solution
Mass of solute = 0.2 x 40 g/mol x 114 g
Mass of solute = 8 g
Therefore, the mass of the non-volatile solute required to reduce the vapour pressure of 114 g of octane to 80% is 8 g. Option (a) is the correct answer.
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