Question

Rectified spirit contains $95\% $ ethanol by weight. The mole fraction of ethanol will be:
A. \[0.881\\ \]
B. \[0.99\\ \]
C. \[0.118\\ \]
D. \[0.81\\ \]

Answer 1

Hint:
Rectified spirit contains $95\% $ ethanol by weight. It means ethanol with $95\% $ purity is known as rectified spirit. For the calculation of the number of moles, we need to divide mass by its molecular mass.

Complete step by step solution
Rectified spirit is highly concentrated alcohol, namely ethanol. It is purified by the process of rectification which is the process of repeated distillation.

Ethanol is an organic compound comprising the hydroxyl functional group. Ethanol is a hydrocarbon with chemical formula ${{\rm{C}}_{\rm{2}}}{{\rm{H}}_{\rm{6}}}{\rm{O}}$ which is written as ${{\rm{C}}_{\rm{2}}}{{\rm{H}}_{\rm{5}}}{\rm{OH}}$. The main characteristics of ethanol are; it is a colourless liquid with a slight characteristic’s odour. Also ethanol is flammable and volatile.
As we know rectified spirit contains $95\% $ ethanol by weight this means if there is ${\rm{100}}\,{\rm{g}}$ of rectified spirit then, the amount of ethanol in it will be ${\rm{95}}\,{\rm{g}}$ and the amount of water will be ${\rm{5}}\,{\rm{g}}$. To calculate the mole fraction of ethanol first of all we have to calculate the number of moles present. This is because the mole fraction is nothing but the quantitative representation of the number of molecules of a particular constituent in a mixture with respect to the total number of moles in the given mixture. So, to find the mole fraction of ethanol in the rectified spirit we have to proceed in the following way;
Amount of ethanol in ${\rm{100}}\,{\rm{g}}$ of spirit is ${\rm{95}}\,{\rm{g}}$.
Amount of water (${H_2}O$) in ${\rm{100}}\,{\rm{g}}$ of spirit is ${\rm{5}}\,{\rm{g}}$.
So, mole fraction of ethanol, (x) is $\dfrac{{{\rm{moles of ethanol}}}}{{{\rm{total moles}}}}$.
Molecular weight of ethanol is $46\;{\rm{g}}\;{\rm{mo}}{{\rm{l}}^{ - 1}}$
Molecular weight of water is $18\;{\rm{g}}\;{\rm{mo}}{{\rm{l}}^{ - 1}}$
First we have to calculate the number of moles,
$Mole\left( n \right) = \dfrac{{mass}}{{molar\,mass}}$
The number of moles of ${{\rm{C}}_{\rm{2}}}{{\rm{H}}_{\rm{5}}}{\rm{OH}}$(${n_{{C_2}{H_5}OH}} $)$ = \dfrac{{95}}{{46}} = 2.06$
The number of moles of ${H_2}O$(${n_{{H_2}O}} $)$ = \dfrac{5}{{18}} = 0.27$
Therefore, mole fraction of ethanol is;
$
 = \dfrac{{{n_{{C_2}{H_5}OH}}}}{{{n_{{H_2}O + }}{n_{{C_2}{H_5}OH}}}}\\
 = \dfrac{{2.06}}{{0.27 + 2.06}}\\
 = 0.881
$

Hence option (A), is the correct option.

Note:
Rectified spirit is the highly concentrated form of ethanol which contains 95% ethanol by weight. The mole fraction of the ethanol in the rectified spirit will be the number of moles of ethanol with respect to the total number of moles in the spirit.