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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.
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Shubham Kumar 6 years, 5 months ago
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