Question

The vapour pressures of ethanol and methanol are 42.0 mm of Hg and 88.5mm of Hg, respectively. An ideal solution is formed at the same temperature by mixing 46.0 gm of ethanol and 16.0 gm of methanol. The mole fraction of methanol vapour is:
A.0.467
B.0.502
C.0.513
D.0.556:

Answer 1

Hint:
Mole Fraction of A in a solution of A and B can be calculated using the formula:
${X_A} = \dfrac{{n.A}}{{n.A + n.B}}$
Where ${X_A}$ = Mole Fraction of A
n.A = No. of Moles of A
n.B = No. of Moles of B

Complete step by step answer:
We are given a solution of ethanol and methanol. Moreover, it is an ideal solution, so we can eliminate the possibility that there is any impurity in the solution that might affect our calculations.
When trying to use the above given formula in our question, substitute A for Methanol and B for methanol.
Hence,
${X_{C{H_3}OH}} = \dfrac{{n.C{H_3}OH}}{{n.C{H_3}OH + n.{C_2}{H_5}OH}}$
Now, since we do not have the values for the number of moles for either Ethanol or Methanol,
We can calculate the number of moles using the following formula:

${\text{No}}{\text{. of Moles of A = }}\dfrac{{{\text{Weight of the given sample of A}}}}{{{\text{Atomic/molecular weight of A}}}}$

Now,
Calculating the molecular weights of ethanol and methanol we get:
mol. weight of methanol = 1(atomic weight of carbon) +4(atomic weight of hydrogen)
+1(atomic weight of oxygen)
= 1(12) +4(1) + 1(16)
= 32 gm/mol

mol. weight of ethanol = 2(atomic weight of carbon) +6(atomic weight of hydrogen)
+1(atomic weight of oxygen)
= 2(12) +6(1) +1(16)
= 46m/mol

Substituting the above derived values in the formula for calculating the Mole Fraction, we get;

$\dfrac{{\dfrac{{16}}{{32}}}}{{\dfrac{{16}}{{32}} + \dfrac{{46}}{{46}}}} = \dfrac{{0.5}}{{1.5}}$
Hence, \[{X_{C{H_3}OH}} = \dfrac{1}{3}\]
In any given solution, when we add the mole fractions of all constituents, we get the sum to turn out as 1.
Using the same theory, we can calculate the mole fraction of ethanol.
i.e., ${X_{{C_2}{H_5}OH}} = \dfrac{2}{3}$
Now, the total pressure can be calculated using the following relation,
where a = ethanol
  b = methanol
${P_{total}} = X{}_a{P_a} + X{}_b{P_b} = 42 \times \dfrac{2}{3} + 88.5 \times \dfrac{1}{3} = 57.5$mm of Hg
So, the mole fraction of methanol vapor = $\dfrac{{{P_b} \times {X_b}}}{{{P_{total}}}} = \dfrac{{88.5 \times \dfrac{1}{3}}}{{57.5}} = 0.513$

Hence, Option C is the correct answer.

Note:
Mole fraction is also known as molar fraction. Mole fraction is temperature independent. When added together the mole fraction of all components of a solution will be equal to 1.