Question

the vapour pressure of two liquids A and B are 50 mm and 100 mm respectively the ratio of mole fractions of B and a in water vapour phase over the liquid if they are mixed in 1 is to 1 ratio is​

Answer 1

Answer:

50:100

Explanation:

50:100=2 this is 2mm

Answer 2

The ratio of mole fractions of B and A in water vapour phase over the liquid if they are mixed in 1 is to 1 ratio is​ $\frac{2}{1}$

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 $p_2=x_2P_2^0$

x = mole fraction

$p^0$= pressure in the pure state

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

$p_{total}=p_1+p_2p_{total}=x_Ap_A^0+x_BP_B^0$

$x_{A}=\frac{\text {moles of A}}{\text {moles of A+moles of B}}=\frac{1}{1+1}=0.5$

$x_{B}=\frac{\text {moles of B}}{\text {moles of A+moles of B}}=\frac{1}{1+1}=0.5$

$p_{A}^0=50 mmHg$

$p_{B}^0=100 mmHg$

$p_{total}=0.5\times 50+0.5\times 100=75mmHg$

$y_{A}$ = mole fraction of A in vapor phase=$y_{A}=\frac{p_{A}}{p_{total}}=\frac{0.5\times 50}{75}=0.33$

$y_{B}=1-y_{A}=1-0.33=0.67$

$\frac{y_{B}}{y_{A}}=\frac{0.67}{0.33}=\frac{2}{1}$

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