So, here's the question which I'm having trouble solving :

Suppose that we condense a gas mixture. At 25 celsius, what will be the pressure of benzene and of toluene over the obtained liquid.

We know that vapor pressures of pure benzene and pure toluene at 25 celsius are 12,7 kPa and 3,79 kPa respectively. We also previously prepared a solution that has equal molar fraction of benzene and toluene. We had to determine the vapor pression of benzene and toluene. (this is an ideal solution)

This gave us: $$p_{benz}=x_{benz} * p^°_{benz}= 0.500 * 12.7 \text{ kPa} = 6,35 \text{ kPa}$$

$$P_{tol}=x_{tol} * p^°_{tol}= 0.500 * 3.79 = 1.90 \text{ kPa}$$ Total pressure: $P = 6.35 + 1.90 = 8.25 \text{kPa}$

Finally, there was another question which asked us what was the composition, in molar fraction, of the vapor in balance with the solution of benzene and toluene at 25 celsius.

This gave us x benz =P benz /Ptotal = 6.35 / 8.25= 0.770

x tol =P tol /P total = 1.90 / 8.25 = 0.230

By the way, here's the answer to my question :

Ptol = Xtol * P^o tol = 0,770 × 3,79 = 2,91 kPa

Pbenz = Xbenz * P^o benz = 0,230 × 12,7 = 2,92 kPa

So, where does the Xtol= 0.770 and Xbenz=0.230 come from ? Thank you

So, here's the question which I'm having trouble solving :

Suppose that we condense a gas mixture. At 25 celsius, what will be the pressure of benzene and of toluene over the obtained liquid.

We know that vapor pressures of pure benzene and pure toluene at 25 celsius are 12,7 kPa and 3,79 kPa respectively. We also previously prepared a solution that has equal molar fraction of benzene and toluene. We had to determine the vapor pression of benzene and toluene. (this is an ideal solution)

This gave us: $$p_{benz}=x_{benz} * p^°_{benz}= 0.500 * 12.7 \text{ kPa} = 6,35 \text{ kPa}$$

$$P_{tol}=x_{tol} * p^°_{tol}= 0.500 * 3.79 = 1.90 \text{ kPa}$$ Total pressure: $P = 6.35 + 1.90 = 8.25 \text{kPa}$

Finally, there was another question which asked us what was the composition, in molar fraction, of the vapor in balance with the solution of benzene and toluene at 25 celsius.

This gave us $$x_{benz} =p_{benz} /P_{total} = 6.35 / 8.25= 0.770$$

$$x_{tol} =p_{tol} /P_{total} = 1.90 / 8.25 = 0.230$$

By the way, here's the answer to my question :

$$p_{tol} = x_{tol} * p^°_{tol} = 0,770 × 3,79 = 2,91 \text{ kPa}$$

$$p_{benz} = x_{benz} * p^°_{tol} = 0,230 × 12,7 = 2,92 \text{ kPa}$$

So, where does the Xtol= 0.770 and Xbenz=0.230 come from ? Thank you

So, here's the question which I'm having trouble solving :

Suppose that we condense a gas mixture. At 25 celsius, what will be the pressure of benzene and of toluene over the obtained liquid.

We know that vapor pressures of pure benzene and pure toluene at 25 celsius are 12,7 kPa and 3,79 kPa respectively. We also previously prepared a solution that has equal molar fraction of benzene and toluene. We had to determine the vapor pression of benzene and toluene. (this is an ideal solution)

This gave us Pbenz=Xbenz * P^o benz= 0.500 * 12.7 kPa = 6,35 kPa

         Ptolz=Xtol * P^o tol= 0.500 * 3.79 = 1.90 kPa

Total pressure= 6.35 + 1.90 = 8.25 kpa

finally, there was another question which asked us what was the composition, in molar fraction, of the vapor in balance with the solution of benzene and toluene at 25 celsius.

This gave us x benz =P benz /Ptotal = 6.35 / 8.25= 0.770

x tol =P tol /P total = 1.90 / 8.25 = 0.230

By the way, here's the answer to my question :

Ptol = Xtol * P^o tol = 0,770 × 3,79 = 2,91 kPa

Pbenz = Xbenz * P^o benz = 0,230 × 12,7 = 2,92 kPa

So, where does the Xtol= 0.770 and Xbenz=0.230 come from ? Thank you

So, here's the question which I'm having trouble solving :

Suppose that we condense a gas mixture. At 25 celsius, what will be the pressure of benzene and of toluene over the obtained liquid.

We know that vapor pressures of pure benzene and pure toluene at 25 celsius are 12,7 kPa and 3,79 kPa respectively. We also previously prepared a solution that has equal molar fraction of benzene and toluene. We had to determine the vapor pression of benzene and toluene. (this is an ideal solution)

This gave us: $$p_{benz}=x_{benz} * p^°_{benz}= 0.500 * 12.7 \text{ kPa} = 6,35 \text{ kPa}$$

$$P_{tol}=x_{tol} * p^°_{tol}= 0.500 * 3.79 = 1.90 \text{ kPa}$$ Total pressure: $P = 6.35 + 1.90 = 8.25 \text{kPa}$

Finally, there was another question which asked us what was the composition, in molar fraction, of the vapor in balance with the solution of benzene and toluene at 25 celsius.

This gave us x benz =P benz /Ptotal = 6.35 / 8.25= 0.770

x tol =P tol /P total = 1.90 / 8.25 = 0.230

By the way, here's the answer to my question :

Ptol = Xtol * P^o tol = 0,770 × 3,79 = 2,91 kPa

Pbenz = Xbenz * P^o benz = 0,230 × 12,7 = 2,92 kPa

So, where does the Xtol= 0.770 and Xbenz=0.230 come from ? Thank you