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Some Basic Concepts Of Chemistry

Question
CBSEENCH11005112

Calculate the molarity of a solution of ethanol in water in which the mole fraction of ethanol is 0.040.(assume the density of water to be one).

Solution
Mole fraction of ethanol is given by
straight C presuperscript straight x subscript 2 straight H subscript 5 OH space equals space fraction numerator straight C presuperscript straight n subscript 2 straight H subscript 5 OH over denominator straight C presuperscript straight n subscript 2 straight H subscript 5 OH space plus straight H presuperscript straight n subscript 2 straight O end fraction space space space space
space space space space space space space space space space space space space equals space 0.040 space space space space space space space space space space space space space space space space space space space space space space space space left square bracket given right square bracket space... left parenthesis 1 right parenthesis
Our aim is to find the number of moles of ethanol in 1L of the solution which is nearly = 1L of water.
Number of moles in 1L of water
                     equals space fraction numerator 1000 straight g over denominator 18 straight g space mol to the power of negative 1 end exponent end fraction space equals space 55.55 space moles
Substituting the values of straight H presuperscript straight n subscript 2 straight O in eq. (1) we have
   fraction numerator straight C presuperscript straight n subscript 2 straight H subscript 5 OH over denominator straight C presuperscript straight n subscript 2 straight H subscript 5 OH plus 55.55 end fraction space equals 0.040
or space space space 0.96 space straight C presuperscript straight n subscript 2 straight H subscript 5 OH space equals space 55.55 space cross times space 0.040
or space space space space space straight C presuperscript straight n subscript 2 straight H subscript 5 OH space equals space fraction numerator 55.55 space cross times space 0.040 over denominator 0.96 end fraction space equals space 2.31 space mol
Hence molarity of the solution = 2.31 M