MHTCET 19 Apr 2026 Shift 1 Paper

Concept:For a weak base in water, the equilibrium constant $K_b$ relates to the initial concentration $C$ and degree of dissociation $\alpha$ by the expression $K_b = \frac{C \alpha^2}{1-\alpha}$.Explanation:Given that the weak base is $1.3\%$ dissociated, we write:$\alpha = \frac{1.3}{100} = 0.013$The given base dissociation constant is:$K_b = 1.69 \times 10^{-5}$Rearrange the formula to solve for concentration $C$:$C = \frac{K_b(1-\alpha)}{\alpha^2}$Substitute the known values:$C = \frac{(1.69 \times 10^{-5})(1-0.013)}{(0.013)^2}$$C = \frac{(1.69 \times 10^{-5})(0.987)}{1.69 \times 10^{-4}}$$C = 0.0987\ \mathrm{M} \approx 0.1\ \mathrm{M}$Thus, the concentration of the weak base solution is approximately $0.1\ \mathrm{M}$.Answer:The concentration of the aqueous solution of the weak base is $0.1\ \mathrm{M}$.Hence, the correct option is B.