Ncert Exercises & Solutions

1.29 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).

Mole fraction of C_{2} H_{5} OH = \frac{Number of moles of C_{2} H_{5} OH}{Number of moles of solution}

0.040 = \frac{n_{c_{2} H_{5} OH}}{n_{c_{2} H_{5} OH} + n_{H_{2} O}} . . . . . . (i)

Number of moles present in 1 L water :

n_{H_{2} O} = \frac{1000 g}{18 g {mol}^{- 1}}

n_{H_{2} O} = 55.55 mol

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

\frac{n_{c_{2} H_{5} OH}}{n_{c_{2} H_{5} OH} + 55.55} = 0.040

n_{c_{2} H_{5} OH} = 0.040 n_{c_{2} H_{5} OH} + (0.404) (55.55)

0.96 n_{c_{2} H_{5} OH} = 2.222 mol

n_{c_{2} H_{5} OH} = \frac{2.222}{0.96} mol

n_{c_{2} H_{5} OH} = 2.314 mol

∴ Molarity of solution = \frac{2.314 mol}{1 L}

= 2.314 M

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