JEE Chemistry 2023 Question with Solution

Given: Mass of hydrated oxalic acid = $3.1500 \, \text{g}$, molar mass = $126 \, \text{g/mol}$, volume of solution = $250.0 \, \text{mL} = 0.250 \, \text{L}$.

Find: Whether Assertion A and Reason R are true, and whether R explains A.

From the solution:

$$M = \frac{\text{moles of solute}}{\text{volume of solution in liters}}$$

First, calculate moles of hydrated oxalic acid:

$$\text{moles} = \frac{3.1500 \, \text{g}}{126 \, \text{g/mol}} = 0.0250 \, \text{mol}$$

Now calculate molarity:

$$M = \frac{0.0250 \, \text{mol}}{0.250 \, \text{L}} = 0.1 \, \text{M}$$

Therefore, Assertion A is true.

The solution also states that the molar mass of hydrated oxalic acid is indeed $126 \, \text{g/mol}$, so Reason R is true.

However, there is a discrepancy in the provided solution: the explanation concludes that R explains A, while the marked option is C. The answer is taken as C.

Therefore, the correct option is C.

Given:

• Mass = $3.1500 \, \text{g}$
• Molar mass = $126 \, \text{g/mol}$
• Volume = $250.0 \, \text{mL} = 0.250 \, \text{L}$

Find: The truth values of Assertion A and Reason R.

Using the molarity relation:

$$M = \frac{n}{V}$$

where $n$ is moles of solute and $V$ is volume in litres.

Step 1: Compute moles.

$$n = \frac{3.1500}{126} = 0.0250 \, \text{mol}$$

Step 2: Compute molarity.

$$M = \frac{0.0250}{0.250} = 0.1 \, \text{M}$$

So Assertion A is true.

Step 3: Check Reason R. The value $126 \, \text{g/mol}$ is exactly the molar mass used in the calculation, so Reason R is true.

provided inconsistency note: the body text supports option D, but the solution marks option C as correct. As instructed, the solution is treated, so the extracted answer is C.