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).

Asked by Abhisek | 1 year ago |  81

1 Answer

Solution :-

Mole fraction of \(C_{ 2 }H_{ 5 }OH\)

=\(\dfrac{Number \; of \; moles \; of \; C_{ 2 }H_{ 5 }OH}{Number \; of \; moles \; of \; solution}\)

0.040 = \(\dfrac{n_{C_{ 2 }H_{ 5 }OH}}{n_{C_{ 2 }H_{ 5 }OH} \; + \; n_{H_{ 2 }O}}\)——(1)

No. of moles present in 1 L water:

\(n_{ H_{ 2 }O} \; = \; \dfrac{ 1000 \; g}{18 \; g \; mol^{ -1 }}[n_{ H_{ 2 }O} = 55.55 mol]\)

Substituting the value of \(n_{ H_{ 2 }O}\) in eq (1),

\(\dfrac{n_{C_{ 2 }H_{ 5 }OH}}{n_{C_{ 2 }H_{ 5 }OH} \; + \; 55.55} = 0.040\)

\(n_{C_{ 2 }H_{ 5 }OH}= 0.040n_{C_{ 2 }H_{ 5 }OH}+ (0.040)(55.55)\)

\( 0.96n_{C_{ 2 }H_{ 5 }OH} = 2.222 mol\)

\(n_{C_{ 2 }H_{ 5 }OH}= \dfrac{ 2.222 }{ 0.96 } \; mol n_{C_{ 2 }H_{ 5 }OH}\)

\( = 2.314 mol\)

Therefore, molarity of solution

=\(\dfrac{ 2.314 \; mol }{ 1 \; L }\)

= 2.314 M

Answered by Pragya Singh | 1 year ago

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