What is the mol fraction of ethylene glycol in the solution phase for an aqueous solution with a vapor pressure of #"760 torr"# if the pure vapor pressure was #"1077 torr"#?

Question

Noah Aldrich · Accepted Answer

I got #0.294#.

As per Raoult's law:

#P_j = chi_j^lP_j^"*"#,

where

There was a drop in vapor pressure because the solution's total vapor pressure—which is less than the pure vapor pressure of water—was disclosed.

I assume however, that the #"1 atm"# #=# #"760 torr"# is for the WATER in the solution, not for the ENTIRE solution (though the question was written in such a way that it seemed to be for the overall solution...).

To monitor that alteration:

#DeltaP = P_i - P_i^"*"#,

where #P_i^"*"# was the vapor pressure of water before adding ethylene glycol, and #P_i# is the vapor pressure of water after adding ethylene glycol.

After that, applying Raoult's law:

#DeltaP = chi_i^lP_i^"*" - P_i^"*"#

#= P_i^"*"(chi_i^l - 1)#

#= -chi_j^lP_i^"*"#

The shift in pressure was:

#"760 torr"# #-# #"1077 torr" = -"317 torr"#

Thus, if ethylene glycol were present in the solution rather than the vapor phase, its mole fraction would be:

#color(blue)(chi_j^l) = -(DeltaP)/(P_i^"*")#

#= -(-"317 torr")/("1077 torr")#

#= color(blue)(0.294)#

Cameron Andrews · Answer

undefined The mole fraction of ethylene glycol in the solution phase can be calculated using Raoult's law, which states that the vapor pressure of a solution is proportional to the mole fraction of each component in the solution. The formula to calculate the mole fraction ($X$) of ethylene glycol in the solution phase is:

$$X = \frac{{P_{\text{{ethylene glycol}}}}}{{P_{\text{{ethylene glycol}}} + P_{\text{{water}}}}$$

Given that the vapor pressure of the solution ($P_{\text{{solution}}}$) is equal to the vapor pressure of water ($P_{\text{{water}}}$) due to Raoult's law, we can set:

$$P_{\text{{solution}}} = P_{\text{{water}}} = 760 \, \text{{torr}}$$

The vapor pressure of pure ethylene glycol ($P_{\text{{ethylene glycol}}}$) is given as $1077 \, \text{{torr}}$. Plugging in these values into the formula for the mole fraction, we can calculate $X$.

Serenity Allen · Answer

undefined To calculate the mole fraction of ethylene glycol ($X_{EG}$) in the solution phase, you can use Raoult's law:

$$P_{solution} = X_{EG} \times P_{EG} + (1 - X_{EG}) \times P_{water}$$

Given:
$$P_{solution} = 760 \text{ torr}$$
$$P_{EG} = 1077 \text{ torr}$$
$$P_{water} = 760 \text{ torr}$$

Rearranging the equation to solve for $X_{EG}$:

$$X_{EG} = \frac{P_{solution} - P_{water}}{P_{EG} - P_{water}}$$

Substituting the given values:

$$X_{EG} = \frac{760 - 760}{1077 - 760}$$

$$X_{EG} = \frac{0}{317}$$

$$X_{EG} = 0$$

So, the mole fraction of ethylene glycol in the solution phase ($X_{EG}$) is 0.