Question

For $30$ mL of $C{H_3}OH$ ( $d = 0.8g/c{m^3}$ ) is mixed with $60$ mL of ${C_2}{H_5}OH$ ( $d = 0.92g/c{m^3}$ ) at ${25^0}C$ to form a solution of density $0.88g/c{m^3}$ . Select the correct option.
(This question has multiple correct options)
A.Molarity and molality of resulting solutions are $6.33$ and $13.59$ respectively.
B.The mole fraction of solute and molality are $0.385$ and $13.59$ respectively.
C.Molarity and $\% $ change in volume are $13.59$ and zero respectively.
D.Mole fraction of solvent and molality are $0.615$ and $13.59$ respectively.

Answer 1

Hint:Molality is the number of moles of solute present in $100g$ of solvent. Molarity is the number of moles of solute present in $1L$ of the solution. Mole fraction is the number of moles of a particular component divided by the total number of moles of all components.

Complete step by step answer:
We have a mixture of two compounds - $C{H_3}OH$ (methanol) and ${C_2}{H_5}OH$ (ethanol). Given that,
Volume of methanol ${V_1} = 30mL$
Density of methanol ${d_1} = 0.8g/c{m^3}$
Volume of ethanol ${V_2} = 60mL$
Density of ethanol ${d_1} = 0.92g/c{m^3}$
Density of resulting solution, ${d_3} = 0.88g/c{m^3}$
Since volume of ethanol is higher than that of methanol, methanol is the solute and ethanol is the solvent.
Weight of methanol present, ${w_1} = {V_1} \times {d_1} = 30 \times 0.8 = 24g$
Weight of ethanol present, ${w_2} = {V_2} \times {d_2} = 60 \times 0.92 = 55.2g$
Volume of resulting solution, ${V_3} = \dfrac{{{w_1} + {w_2}}}{{{d_3}}} = \dfrac{{24 + 55.2}}{{0.88}} = 90mL$
Also when adding the volume of solute and solvent, we get volume of solution as $90mL$. Hence $\% $ change in volume is zero.
Weight of resulting solution ${w_3} = {V_3} \times {d_3} = 90 \times 0.88 = 79.2g$
Molar mass of methanol, ${M_1} = 32g/mol$
Molar mass of ethanol, ${M_2} = 46g/mol$
Number of moles of methanol, ${n_1} = \dfrac{{{w_1}}}{{{M_1}}} = \dfrac{{24}}{{32}} = 0.75$
Number of moles of ethanol, ${n_2} = \dfrac{{{w_2}}}{{{M_2}}} = \dfrac{{55.2}}{{46}} = 1.2$

Molality of the solution can be calculated by the formula,
Molality, $m = \dfrac{{n \times 1000}}{W}$
Where, n is the number of moles of solute and W is the weight of solvent.
Hence molality of solution can be calculated by the formula,
Molality, $m = \dfrac{{{n_1} \times 1000}}{{{w_2}}}$ (because methanol is solute and ethanol is solvent)
Substituting the values,
$m = \dfrac{{0.75 \times 1000}}{{55.2}} = 13.59m$
Similarly molarity of the solution can be calculated by the formula,
Molarity, \[M = \dfrac{{n \times 1000}}{V}\]
Where,
n = number of moles of solute
V = volume of solution.
Hence molality of solution can be calculated by the formula,
Molarity, $M = \dfrac{{{n_1} \times 1000}}{{{V_3}}}$
Substituting the values,
$M = \dfrac{{0.75 \times 1000}}{{90}} = 8.33M$
\[{\text{Mole fraction of methanol = }}\dfrac{{{\text{number of moles of methanol}}}}{{{\text{number of moles of methanol}} + {\text{number of moles of ethanol}}}}\]
Mole fraction of methanol \[ = \dfrac{{{n_1}}}{{{n_1} + {n_2}}} = \dfrac{{0.75}}{{0.75 + 1.2}} = 0.385\]
Mole fraction of ethanol \[ = \dfrac{{{n_2}}}{{{n_1} + {n_2}}} = \dfrac{{1.2}}{{0.75 + 1.2}} = 0.615\]
Hence molality and molarity of solutions are $13.59m$ and $8.33M$ respectively.
Also mole fraction of solute and solvent are $0.385$ and $0.615$ respectively.
Hence Options B and D are correct.

Note:
The sum of mole fractions of all components in a mixture will be equal to one. Hence we can also calculate mole fraction of solvent by calculating the mole fraction of solute first and then subtracting this value from $1$ .