Question

Calculate the molality of a solution containing 20.7 g of potassium carbonate dissolved in 500 mL of solution.

(Assume the density of solution =1 g $m{L}^{-1}$)

• A. 0.333 m
• B. 0.313 m
• C. 0.500 m
• D. 0.121 m

Answer 1

The correct option is B 0.313 m

$\begin{array}{c}\text{Given,}\\ \text{mass of}{K}_{2}C{O}_3=20.7gm\\ \text{and, volume of solution=}500mL\text{or}0.5L\text{.}\\ \text{and, density of solution =}1gm/mL\text{.}\end{array}$

$\begin{array}{c}\text{So, moles of}{K}_{2}C{O}_3\text{dissolved}=\frac{20.7}{138}mol=0.15\text{moles.}\\ \text{Thus, molality of a solution}=\frac{\text{moles of solute}}{\text{Mass of solvent in}kg}\end{array}$

$\begin{array}{c}\text{Here, mass of solute}=20.7gm\\ \text{and, mass of solution=}(500\times 1)gm=500gm\text{.}\\ \therefore \text{mass of solvent}\left(inkg\right)=\frac{500-20.7}{1000}kg\\ \text{Therefore, molality}=\frac{0.15}{0.4793}m=0.313m\end{array}$

Option B is correct answer