# How many mols are there of methanol and ethanol above the solution if the partial vapor pressures of methanol and ethanol are #"2.619 kPa"# and #"4.556 kPa"# respectively?

##### 1 Answer

**Methanol:**

#chi_("MeOH"(v)) = 0.3650#

**Ethanol:**

#chi_("EtOH"(v)) = 0.6350#

As with many other problems about ideal solutions and volatile solutes and solvents, we refer to **Raoult's law** for ideal solutions and **Dalton's law of partial pressures** for ideal gases, respectively:

#P_A = chi_(A(l))P_A^"*"#

#P_(AB) = P_A + P_B# where:

#chi_(A(l))# is the mol fraction of methanol in the solution phase.#B# indicates ethanol.#P_A# is the vapor pressure of methanol above the solution, in the context of the solution.#"*"# indicates that it's for the pure substance.#P_(AB)# is the total pressure above the solution.

It is not entirely obvious, but it's likely we are given the vapor pressures ** in the context of the solution** (rather than the pure substances), because you wrote "partial vapor pressures":

#P_A = "2.619 kPa"#

#P_B = "4.556 kPa"#

(Otherwise, we would not be able to solve for the mol fractions, because we do not know the total pressure.)

In this case then, the **total pressure** from Dalton's law is:

#P_(AB) = P_A + P_B = "7.175 kPa"#

This means we can now solve for the mol fraction of one substance of choice in the vapor phase. Recall that *partial* vapor pressures. So, by considering the definition of partial pressures...

#P_A = chi_(A(v))P_(AB)# where

#(v)# indicates the vapor phase...

...we can then get the mol fractions of each one in the vapor phase.

#=> color(blue)(chi_(A(v))) = P_A/P_(AB) = "2.619 kPa"/"7.175 kPa"#

#= color(blue)(0.3650)#

#=> color(blue)(chi_(B(v))) = 1 - chi_(A(v)) = color(blue)(0.6350)#

However, since mol fractions are always **relative**, without knowing exactly what quantity of one substance was put in, we cannot know the actual

So, apparently, we would have to leave it at knowing the mol fractions...