Question

A solution containing 20 percent acetic acid by mass in water, if the density of the solution at 20$^{0}C$ is 1.2$g/c{{m}^{3}}$. What is the mole fraction of the solute?
a.) 0.069
b.) 0.091
c.) 0.020
d.) 0.3

Answer 1

To solve this question first we should know the basic terminology like mass percent, mole fraction, molarity, molality and the interconversion between these terms.

Complete step by step answer:
Mole fraction – it is defined as the unit of the amount of a constituent expressed in moles divided by the total amount of all the constituents present in the mixture which is also expressed in moles.
Mass percent- mass percent means the mass of the element or the solute divided by the mass of the compound or solute. The final result obtained is multiplied by 100 to convert it into percent.

Given in the question:
20 percent acetic acid by mass
Temperature = 20$^{0}C$
Density of the solution = 1.2 $g/c{{m}^{3}}$
Now, 20 percent acetic acid by mass means 20 g of acetic acid is dissolved in 80 g of water
Density = = 1.2 $g/c{{m}^{3}}$= 1.2$g/ml$
Mass in 1 ml of acetic acid= $\dfrac{x}{1.2} x 100$ = 20
X = 0.24 g
Mass of 1 ml of water = $1.2-0.24=$ 0.96 g
Molar mass of acetic acid i.e. $C{{H}_{3}}COOH$= 60.052
Number of moles of acetic acid will be the ratio of given mass to the molar mass
Number of moles of acetic acid= $\dfrac{0.24}{60.052}$= 0.0039
Molar mass of water i.e. ${{H}_{2}}O$ = 18 g
Number of moles of water = $\dfrac{0.96}{18}=$0.053
Total number of moles = $0.0039+0.0533=$0.572
Mole fraction of acetic acid = $\dfrac{\text{moles of acetic acid}}{\text{total number of moles}}=\dfrac{0.0039}{0.0572}$= 0.068
So, the correct answer is “Option A”.

Note: While doing such a type of question unit conversion is important. For example if we have to calculate molarity and the volume of solution is given in milliliters, we have to convert the volume to liters because molarity is defined as the moles of solutes in per liter of the solution.