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.
If the system of equations \[ x + 2y - 3z = 2, \quad 2x + \lambda y + 5z = 5, \quad 14x + 3y + \mu z = 33 \] has infinitely many solutions, then \( \lambda + \mu \) is equal to:}
The equilibrium constant for decomposition of $ H_2O $ (g) $ H_2O(g) \rightleftharpoons H_2(g) + \frac{1}{2} O_2(g) \quad (\Delta G^\circ = 92.34 \, \text{kJ mol}^{-1}) $ is $ 8.0 \times 10^{-3} $ at 2300 K and total pressure at equilibrium is 1 bar. Under this condition, the degree of dissociation ($ \alpha $) of water is _____ $\times 10^{-2}$ (nearest integer value). [Assume $ \alpha $ is negligible with respect to 1]