Question

If ‘$c$’ is the molarity of a solution, ‘$m$’ the molality, $M_2$ the molecular weight of the solute in a binary solution and ‘$\rho$’, is the density of the solution in $\text{g/cm}^3$, then the relationship between molality ‘$m$’ and molarity ‘$c$’ is given by}

• A. $m = c/[\rho - cM_2/1000]\text{ mol kg}^{-1}$
• B. $m = 1000c/[\rho - cM_2]\text{ mol kg}^{-1}$
• C. $m = c\rho/[1 + cM_2/100]\text{ mol kg}^{-1}$
• D. $m = 1000c/\rho\text{ mol kg}^{-1}$

Hint:

To remember the formula: $\text{Molality} = \frac{1000 \times \text{Molarity}}{1000 \times \text{Density} - \text{Molarity} \times M_{solute}}$. Just ensure all units are consistent.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
Molarity ($c$) is moles of solute per liter of solution. Molality ($m$) is moles of solute per kilogram of solvent. We use the density ($\rho$) to relate solution volume to solution mass, and then subtract the solute mass to find the solvent mass.

Step 2: Detailed Explanation:
Consider $1\text{ L}$ ($1000\text{ cm}^3$) of solution.
1. Moles of solute $= c$.
2. Mass of solute $= c \times M_2$ grams.
3. Mass of solution $= \text{Volume} \times \text{Density} = 1000 \times \rho$ grams.
4. Mass of solvent $= (1000\rho - cM_2)$ grams.
5. Convert solvent mass to kg: $(1000\rho - cM_2)/1000 = \rho - cM_2/1000 \text{ kg}$.
6. Molality $m = \frac{\text{moles of solute}}{\text{mass of solvent in kg}}$:
\[ m = \frac{c}{\rho - cM_2/1000} \]
This matches Option (A). Note that multiplying numerator and denominator by 1000 gives the commonly used form $m = \frac{1000c}{1000\rho - cM_2}$.

Step 3: Final Answer:
The relationship is $m = c/[\rho - cM_2/1000]$.