Q33 / 225 · JEE Main 2004 · Chemistry
The molar solubility product is $\mathrm{K}_{\mathrm{sp}}$. ' $\mathrm{s}$ ' is given in terms of $\mathrm{K}_{\mathrm{sp}}$ by the relation
- A$\mathrm{s}=\left(\frac{\mathrm{K}_{\mathrm{sp}}}{128}\right)^{1 / 4}$
- Correct answerB$\mathrm{s}=\left(\frac{\mathrm{K}_{\mathrm{sp}}}{256}\right)^{1 / 5}$
- C$\mathrm{S}=\left(256 \mathrm{~K}_{\mathrm{sp}}\right)^{1 / 5}$
- D$\mathrm{s}=\left(128 \mathrm{~K}_{\mathrm{sp}}\right)^{1 / 4}$
Solution
$\mathrm{Ksp}=108 s^5$
$1 \times 4^4 \times s^{1+4}=256 s^5=\mathrm{Ksp}$
Hence, the correct answer is option: 'B'.