The mole fraction of methanol (CH3OH) in aqueous solution is 0.10. If density of the solution is 0.97 g mL-1 , the molarity of the solution will be

# The mole fraction of methanol (CH3OH) in aqueous solution is 0.10. If density of the solution is 0.97 g mL-1 , the molarity of the solution will be

1. A

4.0 M

2. B

4.5 M

3. C

5.0 M

4. D

5.5 M

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### Solution:

For x2 = 0.10, let we have n2 = 0.1 mol and n1 = 0.9 mol

The mass of solution will be

$m={n}_{1}{M}_{1}+{n}_{2}{M}_{2}=\left(0.9\mathrm{mol}\right)\left(18\mathrm{g}{\mathrm{mol}}^{-1}\right)+\left(0.1\mathrm{mol}\right)\left(32{\mathrm{mol}}^{-1}\right)=19.4\mathrm{g}$

The volume of solution will be

$V=\frac{m}{\rho }=\frac{19.4\mathrm{g}}{0.97{\mathrm{gmL}}^{-1}}=20.0\mathrm{ml}=20×{10}^{-3}\mathrm{L}$

The molarity of solution will be

$M=\frac{{n}_{2}}{V}=\frac{0.1\mathrm{mol}}{20×{10}^{-3}\mathrm{L}}=5.0{\mathrm{molL}}^{-1}$  