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 per litre of solution. Molality (m) is moles per kilogram of solvent. We use density to link the volume of solution to the mass of solvent.
Step 2: Key Formula or Approach:
Let volume of solution = 1 L (1000 cm{3}).
Mass of solution = \( 1000 \times \rho \).
Moles of solute = c.
Mass of solute = \( c \times M_2 \).
Mass of solvent = Mass of solution - Mass of solute = \( 1000\rho - cM_2 \).
Step 3: Detailed Explanation:
Molality \( m = \frac{\text{moles of solute}}{\text{mass of solvent in kg}} \).
\( m = \frac{c}{(1000\rho - cM_2)/1000} \).
\( m = \frac{c}{\rho - cM_2/1000} \).
This matches Option A. Alternatively, it is often written as \( m = \frac{1000c}{1000\rho - cM_2} \).
Step 4: Final Answer:
The correct relation is Option A.