Question

At 25°C, the vapour pressure of pure liquid A (molecular weight = 40) is 100 torr while that of pure liquid B (molecular weight = 80) is 40 torr. The vapour pressure at 25°C of a solution containing 20g of each A and B is
(A) 80 torr
(B) 59.8 torr
(C) 68 torr
(D) 48 torr

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Answer 1

Answer :–

Correct answer - Option (A) 80 torr

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Answer 2

The vapour pressure at 25°C of a solution containing 20g of each A and B is 80 torr

Explanation:

According to Raoult's law, the vapor pressure of a component at a given temperature is equal to the mole fraction of that component multiplied by the vapor pressure of that component in the pure state.

$p_1=x_1p_1^0$ and $p_2=x_2P_2^0$

where, x = mole fraction

$p^0$ = pressure in the pure state

According to Dalton's law, the total pressure is the sum of individual pressures.

$p_{total}=p_1+p_2$$p_{total}=x_Ap_A^0+x_BP_B^0$

Moles of A =$\frac{\text{ given mass of A}}{\text{ molar mass of A}}= \frac{20g}{40g/mole}=0.5moles$

Moles of B=$\frac{\text{ given mass of B}}{\text{ molar mass of B}}= \frac{20g}{80g/mole}=0.25moles$

$x_{A}=\frac{\text {moles of A}}{\text {moles of A+moles of B}}=\frac{0.5}{0.5+0.25}=0.67$,  

$x_{B}=\frac{\text {moles of B}}{\text {moles of A+moles of B}}=\frac{0.25}{0.5+0.25}=0.33$,  

$p_{A}^0=100torr$

$p_{B}^0=40torr$

$p_{total}=0.67\times 100+0.33\times 40=80torr$

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