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100 g of liquid A (molar mass 140 g mol^−1) was dissolved in 1000 g of liquid B (molar mass 180 g mol^−1). The vapour pressure of pure liquid B was found to be 500 torr. - CBSE (Science) Class 12 - Chemistry

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Question

100 g of liquid A (molar mass 140 g mol−1) was dissolved in 1000 g of liquid B (molar mass 180 g mol−1). The vapour pressure of pure liquid B was found to be 500 torr. Calculate the vapour pressure of pure liquid A and its vapour pressure in the solution if the total vapour pressure of the solution is 475 Torr.

Solution

Number of moles of liquid A, `n_A = 100/140 "mol"`

= 0.714 mol

Number of moles of liquid B, `n_B = 1000/180 "mol"`

= 5.556 mol

Then, mole fraction of A, `x_A = n_A/(n_A+n_B)`

`= 0.714/(0.714+5.556)`

= 0.114

And, mole fraction of B, xB = 1 − 0.114

= 0.886

Vapour pressure of pure liquid B, `p_B^0`= 500 torr

Therefore, vapour pressure of liquid B in the solution,

`p_B= p_B^0x_B`

= 500 × 0.886

= 443 torr

Total vapour pressure of the solution, ptotal = 475 torr

∴ Vapour pressure of liquid A in the solution,

pA = ptotal − pB

= 475 − 443

= 32 torr

Now,

`p_A = p_A^0x_A`

`=>p_A^0 = p_A/x_A`

`= 32/0.114`

= 280.7 torr

Hence, the vapour pressure of pure liquid A is 280.7 torr.

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Solution 100 g of liquid A (molar mass 140 g mol^−1) was dissolved in 1000 g of liquid B (molar mass 180 g mol^−1). The vapour pressure of pure liquid B was found to be 500 torr. Concept: Ideal and Non-ideal Solutions.
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