Question

A mixture of ethanol and ${\text{CC}}{{\text{l}}_{\text{4}}}$ has 30 % ${\text{CC}}{{\text{l}}_{\text{4}}}$ by weight, What is mole fraction of ${\text{CC}}{{\text{l}}_{\text{4}}}$ in the mixture?
(A) 0.11
(B) 0.89
(C) 0.25
(D) 0.67

Answer 1

We can define the mole fraction of a component as a ratio of the mole of that component to the total number of moles present in the system.
The mole fraction formula in any solution, the mole fraction of solute $A$ is = $\dfrac {Moles\;of\;A}{Total\;number\;of\;moles}$
Mathematically, we can write,
${\text{Mole fraction of A = }}\dfrac{{{\text{Number of moles of A}}}}{{{\text{Total number of moles}}}}{\text{ = }}\dfrac{{{{\text{n}}_A}}}{{{{\text{n}}_{total}}}}$.
Where,
{{\text{n}}_A}$$$$ = Number of moles of A component.
{{\text{n}}_{total}}$$$$ = Total number of moles present in the system.

Complete step by step answer:
Given, A mixture of ethanol and ${\text{CC}}{{\text{l}}_{\text{4}}}$ has 30 % ${\text{CC}}{{\text{l}}_{\text{4}}}$ by weight.
So, we can say, $30g$ ${\text{CC}}{{\text{l}}_{\text{4}}}$ is present in the 100g mixture.
So, Mass of {\text{CC}}{{\text{l}}_{\text{4}}}$$$ = {\text{ }}30g$$ Mass of ethanol = 100 - 30 = 70{\text{gm}}$$\therefore $Moles of${\text{CC}}{{\text{l}}{\text{4}}}$present$ = \dfrac{{30}}{{154}} = 0.195$$\left[ {{\text{Molecular weight of CC}}{{\text{l}}{\text{4}}} = 12 + \left( {35.5 \times 4} \right) = 154} \right]$$\therefore $Moles of${\text{ C}}{{\text{H}}{\text{3}}}{\text{C}}{{\text{H}}{\text{2}}}{\text{OH}};{\text{ = }}\dfrac{{{\text{70}}}}{{{\text{46}}}}{\text{ = 1}}{\text{.521}}$

\left[ {{\text{Molecular weight of Ethanol (}}{{\text{C}}_{\text{2}}}{{\text{H}}_{\text{5}}}{\text{OH) = }}\left( {{\text{12 \times2}}} \right){\text{ + }}\left( {{\text{6\times 1}}} \right){\text{ + 16 = 46}}} \right]
Now, we have the number of moles of both ${\text{CC}}{{\text{l}}_{\text{4}}}$and Ethanol.
So, we can use the formula,
${\text{Mole fraction of CC}}{{\text{l}}_{\text{4}}}{\text{ = }}\dfrac{{{\text{Number of moles of CC}}{{\text{l}}_{\text{4}}}}}{{{\text{Total number of moles}}}}{\text{ = }}\dfrac{{{{\text{n}}_{{\text{CC}}{{\text{l}}_{\text{4}}}}}}}{{{{\text{n}}_{{\text{total}}}}}}$
$\therefore$Mole fraction of ${\text{CC}}{{\text{l}}_{\text{4}}} = \dfrac{{0.195}}{{\left( {0.195 + 1.521} \right)}} = 0.11$

Hence, the correct option (A) is the right answer.

Note: The sum of all the mole fractions is equal to $1$:
$\sum\limits_{i = 1}^N {{n_i} = {n_{tot}};\;\;\sum\limits_{i = 1}^N {{x_i} = 1.} }$
Mole fraction is a quantity which is basically used to define concentration of a component in a mixture.
We can use this quantity to determine many properties like entropy change of mixing, free energy change of mixing, internal energy change of a gaseous system etc.
Mole fraction is a unit-less quantity.