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].

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Solution

We have,

The mole fraction of ethanol is $0.040$

Mole fraction of

$C_{2} H_{5} O H = \frac{n o o f m o l e s o f C_{2} H_{5} O H}{n o o f m o l o f s o l}$

$0.040 = \frac{n C_{2} H_{5} O H}{n C_{2} H_{5} O H + n H_{2} O} . . . . (1)$

Number of moles of water present in 1L

$n H_{2} O = \frac{1000 g}{18 g / m o l}$

Substituting this value in eq1

$\Rightarrow \frac{n C_{2} H_{5} O H}{n C_{2} H_{5} O H + 55.55} = 0.040$

$\Rightarrow C_{2} H_{5} O H = 0.040 (C_{2} H_{5} O) + 0.040 (55.55)$

$\Rightarrow 0.96 C_{2} H_{5} O H = 2.222 m o l$

$n C_{2} H_{5} O H = 2.314 m o l$

Thus, Molarity of the solution

$= \frac{2.314 m o l}{1 L}$

$= 2.314 m / L$

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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).

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