Question

The mole fraction of $C{{H}_{3}}OH$ in an aqueous solution is 0.02 and its density is $0.994g/c{{m}^{3}}$ . Determine its molarity and molality.

Answer 1

The molarity of the solution can be calculated by dividing the number of moles with the volume of the solution. The molality of the solution is calculated by dividing the number of moles of the solute with the mass of the solvent in kg.

Complete step by step answer:
Molarity: The molarity of the solution is defined as the number of moles of the solute present per liter. It is represented by the symbol, M.
$Molarity=\frac{\text{Moles of the solute}}{\text{Volume of the solution}}$
Moles of the solute can be calculated by:
$Moles=\frac{\text{mass of the solute}}{\text{molar mass of the solute}}$
Molality: The molality of a solution is defined as the number of moles of the solute dissolved in 1kg (1000 g) of the solvent. It is represented by the symbol, ‘m’.
$Molality=\frac{\text{Moles of the solute}}{\text{Mass of the solvent in kg}}$
Now, according to the question: the mole fraction of $C{{H}_{3}}OH$ in an aqueous solution is 0.02.
So, $0.02=\frac{x}{y+x}$
Where x = moles of $C{{H}_{3}}OH$ and y = moles of water.
Moles of water = $\frac{1000\text{ g}}{18}=55.55$
Hence,
$0.02=\frac{x}{55.55+x}$
$x-0.02x=55.55\text{ x 0}\text{.02}$
$\text{0}\text{.98}x=1.111$
$x=1.13\text{ moles}$
So, the moles of solute ($C{{H}_{3}}OH$ ) is 1.13 and mass of solvent is 1kg.
Therefore, the molality of the solution is $\frac{1.13}{1}=1.13\text{ m}$
Now, for molarity of the solution,
The moles of the solute ($C{{H}_{3}}OH$ ) is 1.13.
The volume of the solution can be calculated by dividing the total mass of the solution with the density. Given density = $0.994g/c{{m}^{3}}$
Total mass = molecular mass of $C{{H}_{3}}OH$ x moles of$C{{H}_{3}}OH$ + molecular mass of water x moles of water
Total mass = $32\text{ x 1}\text{.13 + 18 x 55}\text{.55 = 1036}\text{.06}$
Volume = $\frac{1036.06}{0.994}mL=\frac{1036.06}{994}L=1.04L$
Now, putting these in molarity equation,
Molarity = $\frac{1.13}{1.04}=1.08M$

Therefore, the molarity of the solution is 1.08 M and the molality of the solution is 1.13 m.

Note: The volume of the solution must be taken in liters only. The value of a mole fraction of water can directly be taken as 55.55 because the aqueous solution will have water as the solvent.