JEE Main 2024 — Some Basic Concepts Of Chemistry Question with Solution

To find the molality of the solution, we need to follow these steps:

1. Determine the number of moles of anhydrous $${\mathrm{CuSO}}_4$$ in the solution.

The molarity (M) is given as $$2\times 10^{-1}$$ M in a 500 mL solution. Hence, the number of moles of $${\mathrm{CuSO}}_4$$ is:

$$\text{Moles of }{\mathrm{CuSO}}_4=\text{Molarity}\times \text{Volume in liters}=2\times 10^{-1}\times 0.5=0.1\text{ moles}$$

2. Calculate the mass of the solution using the given density.

The density of the solution is given as $$1.25g/mL$$. The volume of the solution is 500 mL. Therefore, the mass of the solution is:

$$\text{Mass of solution}=\text{Density}\times \text{Volume}=1.25\times 500=625\text{ grams}$$

3. Find the mass of the solvent (water) in the solution by subtracting the mass of the solute (anhydrous $${\mathrm{CuSO}}_4$$) from the total mass of the solution.

We know the number of moles of $${\mathrm{CuSO}}_4$$, and we can find its molar mass.

$$\text{Molar mass of }{\mathrm{CuSO}}_4=63.5+32+4\times 16=159.5\text{ g/mol}$$

Therefore, the mass of $${\mathrm{CuSO}}_4$$ is:

$$\text{Mass of }{\mathrm{CuSO}}_4=\text{Number of moles}\times \text{Molar mass}=0.1\times 159.5=15.95\text{ grams}$$

4. Calculate the mass of the solvent (water):

$$\text{Mass of water}=\text{Mass of solution}-\text{Mass of }{\mathrm{CuSO}}_4=625-15.95=609.05\text{ grams}$$

Convert the mass of water to kilograms:

$$\text{Mass of water}=609.05\text{ grams}=0.60905\text{ kilograms}$$

5. Calculate the molality using the formula:

$$\text{Molality}(\text{m})=\frac{\text{Moles of solute}}{\text{Mass of solvent in kg}}=\frac{0.1}{0.60905}=0.164\text{ mol/kg}$$

Converting to the desired unit:

$$\text{Molality}(\text{m})=0.164\times 10^3=164\times 10^{-3}$$

Therefore, the molality of the solution is approximately 164 $$\times 10^{-3}\mathrm{m}$$.