Benzene and toluene form ideal solution over the entire range of composition. The vapour pressure of pure benzene and toluene at 300 K are 50.71 mm Hg and 32.06 mm Hg respectively. Calculate the mole fraction of benzene in vapour phase if 80 g of benzene is mixed with 100 g of toluene.
Molar mass of benzene(C6H6) = 6 X 12 + 6 X 1 = 78 g/mol
Molar mass of toluene = 7 x 12 + 8 x 1 = 92 g/mol
Now no of moles in 80g of benezen = 80 / 78 = 1.026 mol
No of moles in 100g of toluene = 100 / 92 = 1.087 mol
∴Mole fraction of benzene xb = 1.026 / 1.026 + 1.087 = 0.486
And Mole fraction of toluene,xt = 1 - 0.486 = 0.514
We have given that
Vapor pressure of pure benzene pb° = 50.71 mm Hg
And, vapour pressure of pure toluene, pt° = 32.06 mm Hg
Therefore partial Vapor pressure of benzene, pb = pb X xb
= 50.71 x 0.486
= 24.65 mm Hg
And partial Vapor pressure of toluene, pt = pt X xt
Pt = p°t X xt = 32.06 x 0.514
= 16.48
Total vapour pressure = 24.65 + 16.48 = 41.13 mm Hg
Mole fraction of benzene in vapour phase = 24.65 / 41.13 = 0.60
The reaction between A and B is first order with respect to A and zero order with respect to B. Fill in the blanks in the following table:
| Experiment |
A/ mol L - 1 |
B/ mol L - 1 |
Initial rate/mol L - 1 min - 1 |
| I | 0.1 | 0.1 |
2.0 × 10 - 2 |
| II | -- | 0.2 |
4.0 × 10 - 2 |
| III | 0.4 | 0.4 | -- |
| IV | -- | 0.2 |
2.0 × 10 - 2 |
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Molar mass of benzene(C6H6) = 6 X 12 + 6 X 1 = 78 g/mol
Molar mass of toluene = 7 x 12 + 8 x 1 = 92 g/mol
Now no of moles in 80g of benezen = 80 / 78 = 1.026 mol
No of moles in 100g of toluene = 100 / 92 = 1.087...
Step-by-step explanation:
• Molar mass of benzene(C6H6) = 6 X 12 + 6 X 1 = 78 g/mol
•
• Molar mass of toluene = 7 x 12 + 8 x 1 = 92 g/mol
•
• Now no of moles in 80g of benezen = 80 / 78 = 1
Students often lose marks by skipping steps, writing incomplete explanations, or misunderstanding keywords used in the question.
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