Calculate molarity of solution of ethanol …

Let the mole fraction  of ethanol (C₂H₅OH) be represented by x,
Then the following expression represents its mole fraction in ethanol solution- 
^xC₂h₅OH = {tex}\frac{\mathrm{n}\left(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}\right)}{\mathrm{n}\left(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}\right)+\mathrm{n}\left(\mathrm{H}_{2} \mathrm{O}\right)}{/tex}
(where x C₂H₅OH  represents mole fraction of ethanol)
= 0.040 (given)....(i);
where "n'' (represents number of moles of the two referred liquids in ethanol solution).
The following steps are used for calculations,
The aim is to find number of moles of ethanol in 1 L of the solution.
Step 1 /-  No. of moles in 1 L of water
= {tex}\frac{1000 \mathrm \ {g}}{18 \mathrm \ {g} \mathrm \ {mol}^{-1}}{/tex}
= 55.55 moles
Step 2 /-. Substituting n (H₂O)
= 55.55 in equation (i), we get
x ( C₂H₅OH)
= {tex}\frac{\mathrm{n}\left(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}\right)}{\mathrm{n}\left(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}\right)+55.55}{/tex}
= 0.040......(given )
solving for n (C₂H₅OH) we get,
0.96 n ( C₂ H₅ OH ) 
= 55.55 {tex}\times{/tex} 0.040
{tex}\therefore{/tex} n (C₂H₅OH)
= 2.31 mols
So  2.31 moles of C₂ H₅ OH are present per litre of given ethanol solution. 
Hence,
the molarity of the solution of ethanol is 2.31 M.