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प्रश्न
Heptane and octane form an ideal solution. At 373 K, the vapour pressures of the two liquid components are 105.2 kPa and 46.8 kPa respectively. What will be the vapour pressure of a mixture of 26.0 g of heptane and 35 g of octane?
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उत्तर १
Vapour pressure of heptane `(p_1^circ)` = 105.2 kPa
Vapour pressure of octane `(p_2^circ)` = 46.8 kPa
We know that,
Molar mass of heptane (C7H16) = 7 × 12 + 16 × 1
= 100 g mol−1
∴ Number of moles of heptane = `26/100` mol
= 0.26 mol
Molar mass of octane (C8H18) = 8 × 12 + 18 × 1
= 114 g mol−1
∴ Number of moles of octane = `35/114` mol
= 0.31 mol
Mole fraction of heptane (x1) = `0.26/(0.26 + 0.31)`
= 0.456
And, mole fraction of octane (x2) = 1 − 0.456
= 0.544
Now, partial pressure of heptane (p1) = `x_1 p_1^circ`
= 0.456 × 105.2
= 47.97 kPa
Partial pressure of octane (p2) = `x_2 p_2^circ`
= 0.544 × 46.8
= 25.46 kPa
Hence, the vapour pressure of the solution (ptotal) = p1 + p2
= 47.97 + 25.46
= 73.43 kPa
उत्तर २
Vapour pressure of heptane `(p_1^circ)` = 105.2 kPa
Vapour pressure of octane `(p_2^circ)` = 46.8 kPa
No. of moles of heptane = `26/100` ..(Mol. mass of heptane = 100 g mol−1)
= 0.260
No. of moles of octane = `35/114` ...(Mol. mass of octane = 114 g mol−1)
= 0.307
∴ Total no. of moles in solution = 0.260 + 0.307
= 0.567
Hence, mole fraction of heptane = `0.260/0.567`
= 0.458
Mole fraction of octane = `0.307/0.567`
= 0.541
Since the solution is ideal,
p = `p_"heptane" + p_"octane"`
= (0.458 × 105.2) + (0.541 × 46.8) kpa
= 48.1816 + 25.3188
= 73.50 kpa
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