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

##### 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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