Question

What would be the molality of 20% (mass/mass) aqueous solution of $KI$?
(molar mass of $KI$= 166 g/mol).
A. 1.08
B. 1.48
C. 1.51
D. 1.35

Answer 1

Try to recall that mass percent of a solute in a solution is the mass of the solute in grams present in 100 g of the solution. Now by using this you can easily find the correct option from the given ones.

Complete step by step solution:
It is known to you that molality of a solution is defined as the number of moles of solute dissolved in 1kg of the solvent, i.e., $m = \dfrac{{{\text{number of moles of solute}}}}{{weight{\text{ }}of{\text{ }}solvent{\text{ in kg}}}}$.
Calculation:
20% (mass/mass) aqueous solution of $KI$ means 20 g of $KI$ in 100 g of solution.
Mass of solution= 100g
Mass of solute = 20g
Mass of solvent= mass of solution – mass of solute
$\begin{gathered} = 100 - 20 \\\ = 80g = 0.08kg \\\ \end{gathered}$
Molar mass of $KI$= 166 g/mol

Number of moles of solute, $n = \dfrac{{mass{\text{ of KI}}}}{{molar{\text{ mass of KI}}}}$
$n = \dfrac{{20}}{{166}} = 0.121$.
So, molality of solution, $m = \dfrac{{{\text{number of moles of solute}}}}{{weight{\text{ }}of{\text{ }}solvent{\text{ in kg}}}}$
$\Rightarrow m = \dfrac{{0.121}}{{0.08}} = 1.51$.

Therefore, from the above calculations we can conclude that option C is the correct option to the given question.

Note: It should be remembered to you that molality, mole fraction etc. are preferred over molality, normality etc. This is because the former involve masses of the solute and the solvent which do not change with temperature whereas the latter involve volumes of the solutions which change with temperature.