Question

A $5.2$ molal aqueous solution of methyl alcohol, ${\text{C}}{{\text{H}}_{\text{3}}}{\text{OH}}$is supplied. What is the mole fraction of methyl alcohol in the solution?
(A)- $0.05$
(B)- $0.10$
(C)- $0.18$
(D)- $0.086$

Answer 1

Hint: For calculating the mole fraction of methyl alcohol present in the solution we have to divide the no. of moles of methyl alcohol present in the solution by the total moles of solvent or constituents present in the solution.

Complete answer:
It is given that, molality of aqueous solution of methyl alcohol is $5.2$.
-So, by the definition of molality it means that $5.2$ moles of methyl alcohol present in the${\text{1000Kg}}$ or ${\text{1000g}}$ of water.
-Mole fraction of the methyl alcohol in the aqueous solution is calculated as follow:
     \[Mole\]\[fraction = \dfrac{{{n_{methylalcohol}}}}{{{n_{methylalcohol}} + {n_{water}}}}\]………..(i)
-For calculating the mole fraction of methyl alcohol first, we have to know about the moles of water and that will be determined by the following formula:
$n = \dfrac{W}{M}$…………..(ii)
Where, n = no. of moles,
W = given weight,
M = molecular or molar weight.
So, the molecular weight of water is ${\text{18g/mol}}$ and the given weight is ${\text{1000g}}$. Now putting these values on the equation (ii) we get,
${\text{n = }}\dfrac{{{\text{1000}}}}{{{\text{18}}}}{\text{ = 55}}{\text{.56mol}}$
Hence, no moles of water present in the solution is ${\text{55}}{\text{.6mol}}$.
-Now, we putting values on equation (i) to calculate mole fraction of ${\text{C}}{{\text{H}}_{\text{3}}}{\text{OH}}$
\[Mole\]\[fraction = \dfrac{{5.2}}{{5.2 + 55.6}} = 0.086\]
So, mole fraction of methyl alcohol in the solution is $0.086$.

Hence, option (D) is correct.

Note:
In this question some of you may get confused in between the terms, no. of moles $\left( n \right)$& mole fraction $\left( {{n_i}} \right)$ but both are different and have different formulae to calculate. So, don’t think these two things are similar.