30 mL of CH₃OH (d = 0.8 g/cm3) is mixed with 60 mL of C₂H₅OH (d = 0.92 g/cm³) at 25°C to form a solution of density 0.88 g/cm³.
Select the correct option(s) :

Difficult

A

Molarity and molality of resulting solution are 6.33 M and 13.59 m respectively.
B

The mole fraction of solute and molality are 0.38 and 13.59 m respectively.
C

Molarity and percentage change in volume are 13.5 M and zero respectively.
D

Mole fraction of solvent and molality are 0.62 and 13.59 m respectively.

Solution

30 mL of CH₃OH ............
CH₃OH is solute (less amount)
Mass of CH3OH = 30 × 0.8 = 24 g
Mass of C₂H₅OH = 60 × 0.92 = 55.2 g
Mass of solution = 79.2 g
$\begin{array}{l} Volume of solution = \frac{79.2}{0.88} = 90 mL \\ Molarity = \frac{n_{{CH}_{0} O}}{V (L)} = \frac{24 / 32}{90} \times 1000 = 8.33 M \\ Molality = \frac{24 / 32}{55.2} \times 1000 = 13.59 m \\ Mole fraction of solute = \frac{\frac{24}{32}}{\frac{24}{32} + \frac{55.2}{46}} = 0.38 \\ Mole fraction of solvent = 1 - 0.38 = 0.62 \end{array}$

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