Step 1: Verifying the Assertion A
Given mass of hydrated oxalic acid = 3.1500 g
Molar mass of hydrated oxalic acid = 126 g/mol
Volume of solution = 250.0 mL = 0.250 L
To calculate molarity (M), we use the formula: \[ M = \frac{\text{moles of solute}}{\text{volume of solution in liters}} \] Moles of solute: \[ \text{moles} = \frac{3.1500 \, \text{g}}{126 \, \text{g/mol}} = 0.0250 \, \text{mol} \] Thus, molarity: \[ M = \frac{0.0250 \, \text{mol}}{0.250 \, \text{L}} = 0.1 \, \text{M} \] So, Assertion A is correct.
Step 2: Verifying the Reason R
The molar mass of hydrated oxalic acid is indeed 126 g/mol, as given in the question. This confirms that Reason R is correct.
Thus, both Assertion A and Reason R are true, and Reason R explains Assertion A.
Let \( y = f(x) \) be the solution of the differential equation
\[ \frac{dy}{dx} + 3y \tan^2 x + 3y = \sec^2 x \]
such that \( f(0) = \frac{e^3}{3} + 1 \), then \( f\left( \frac{\pi}{4} \right) \) is equal to:
Find the IUPAC name of the compound.
If \( \lim_{x \to 0} \left( \frac{\tan x}{x} \right)^{\frac{1}{x^2}} = p \), then \( 96 \ln p \) is: 32