Question

Calculate the molality, molarity and mole fraction of potassium iodide if the density of $20\%$ equipotential iodide is 1.202 g/ml.

Answer 1

Hint: $20\%$ equipotential iodide means $20 \ g$ of potassium iodide present in $100 \ g$ of the solution. Use the density to find out the volume of solution and then molarity can be calculated. Find the number of moles of both solute and solvent and you can find out the mole fraction. The formula used for the same is:
\[No.\text{ of moles}=\dfrac{Given\operatorname{mass}}{Molar\operatorname{mass}}\]
Molality can be calculated by taking the ratio of number of moles of solute present in 1kg of solvent.

Complete step-by-step answer:
As it is mentioned in the question, we have 20 g of potassium iodide in 100 g of the solution. Upon adding the individual masses of the atoms of potassium and iodine, we get the molar mass of KI to be 166 g.
We can find the number of moles of KI by using the following formula-
\[No.\text{ of moles}=\dfrac{Given\operatorname{mass}}{Molar\operatorname{mass}}=\dfrac{20}{166}=0.12\operatorname{moles}\]
Molarity is defined as the number of moles of solute present in 1000 ml of the solution. We need the density of the solution to find out the volume of solution. This is done as follows-
\[\begin{align}
  & 1.202\operatorname{g}\text{of potassium iodide solution}\to 1\operatorname{ml} \\
 & 100\operatorname{g}\to \dfrac{1}{1.202}\times 100=83.194\operatorname{ml} \\
\end{align}\]
So, 83.194 ml of solution contains 0.12 moles of KI.
\[1000\operatorname{ml}\text{ of solution contains}\to \dfrac{0.12}{83.194}\times 1000=1.442\operatorname{M}\]
So, the given solution is 1.442 Molar.
We have already mentioned above that 20 g of potassium iodide is present in 100 g of solution, which also means that 0.12 moles of KI is present in 80 g of solvent. Molality of the solution is defined as the number of moles of solute present in 1000 g of solvent. Below, we calculate molality using unitary method as-
\[\begin{align}
  & 80\operatorname{g}\text{of solvent}\to 0.12\operatorname{g}\text{of KI} \\
 & \text{1000}\operatorname{g}\text{of solvent}\to \dfrac{0.12}{80}\times 1000=1.5\operatorname{m} \\
\end{align}\]
So, the given solution is 1.5 molal.
Mole fraction is defined as the ratio between the number of moles of one constituent with the total number of moles present in the solution. We already have the number of moles of solute (which is potassium iodide) as 0.12 moles. We have to find the number of moles of 80 g of solvent present in the solution. Now, the molar mass of water is 18 g. Therefore,
\[No.\text{ of moles of water}=\dfrac{Given\operatorname{mass}}{Molar\operatorname{mass}}=\dfrac{80}{18}=4.44\operatorname{moles}\]
So,
\[\text{Mole fraction of KI}=\dfrac{0.12}{0.12+4.44}=0.0263\]
We get the mole fraction of KI to be 0.0263.

Note: All the stoichiometric calculations done above are very unit specific. A student is therefore advised caution during performing calculations. The mole fraction is a dimensionless quantity as it is a ratio between the same quantities, i.e. moles. The density given here in the question is the density of the overall solution and not of any particular constituent.