MHT CET 2026 16th April Morning Shift | Ionic Equilibrium Question 27 | Chemistry

The solubility product of a sparingly soluble salt BA is $6.4 \times 10^{-13}$. Calculate it's solubility in $\text{g dm}^{-3}$ .
Molar mass of salt is $190\ \text{g mol}^{-1}$ .

$1.52 \times 10^{-4}$

$1.25 \times 10^{-4}$

$2.1 \times 10^{-4}$

$1.9 \times 10^{-4}$

MHT CET 2026 16th April Morning Shift

MCQ (Single Correct Answer)

-0

The solubility of AgCl in 0.1 M NaCl is S mol/L. If the solubility product of AgCl is $1.8 \times 10^{-10}$, then S is approximately:

$1.8 \times 10^{-9}$ M

$1.8 \times 10^{-10}$ M

$1.8 \times 10^{-11}$ M

$1.8 \times 10^{-12}$ M

MHT CET 2026 16th April Morning Shift

MCQ (Single Correct Answer)

-0

Identify from following a weak electrolyte based on their dissociation nature in aqueous medium.

$\text{NH}_3$

$\text{NaOH}$

$\text{KOH}$

$\text{HI}$

MHT CET 2026 15th April Evening Shift

MCQ (Single Correct Answer)

-0

What is the solubility product of binary sparingly soluble salt BA if 100 mL saturated solution of salt consist $10^{-4}$ moles at room temperature?

$1 \times 10^{-4}$

$1 \times 10^{-6}$

$1 \times 10^{-8}$

$1 \times 10^{-10}$

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