1
JEE Main 2025 (Online) 24th January Morning Shift
MCQ (Single Correct Answer)
+4
-1
Change Language

$\mathrm{K}_{\mathrm{sp}}$ for $\mathrm{Cr}(\mathrm{OH})_3$ is $1.6 \times 10^{-30}$. What is the molar solubility of this salt in water?

A
$\sqrt[5]{1.8 \times 10^{-30}}$
B
$\frac{1.8 \times 10^{-30}}{27}$
C
$\sqrt[4]{\frac{1.6 \times 10^{-30}}{27}}$
D
$\sqrt[2]{1.6 \times 10^{-30}}$
2
JEE Main 2025 (Online) 23rd January Evening Shift
MCQ (Single Correct Answer)
+4
-1
Change Language

pH of water is 7 at $25^{\circ} \mathrm{C}$. If water is heated to $80^{\circ} \mathrm{C}$., it's pH will :

A
Decrease
B
Remains the same
C
Increase
D
$\mathrm{H}^{+}$concentration increases, $\mathrm{OH}^{-}$concentration decreases
3
JEE Main 2025 (Online) 23rd January Morning Shift
MCQ (Single Correct Answer)
+4
-1
Change Language

Which of the following happens when $\mathrm{NH}_4 \mathrm{OH}$ is added gradually to the solution containing 1 M $\mathrm{A}^{2+}$ and $1 \mathrm{M} \mathrm{B}^{3+}$ ions?

Given : $\mathrm{K}_{\text {sp }}\left[\mathrm{A}(\mathrm{OH})_2\right]=9 \times 10^{-10}$ and $\mathrm{K}_{\mathrm{sp}}\left[\mathrm{B}(\mathrm{OH})_3\right]=27 \times 10^{-18}$ at 298 K.

A
$\mathrm{A}(\mathrm{OH})_2$ will precipitate before $\mathrm{B}(\mathrm{OH})_3$
B
$\mathrm{A}(\mathrm{OH})_2$ and $\mathrm{B}(\mathrm{OH})_3$ will precipitate together
C
Both $\mathrm{A}(\mathrm{OH})_2$ and $\mathrm{B}(\mathrm{OH})_3$ do not show precipitation with $\mathrm{NH}_4 \mathrm{OH}$
D
$\mathrm{B}(\mathrm{OH})_3$ will precipitate before $\mathrm{A}(\mathrm{OH})_2$
4
JEE Main 2025 (Online) 22nd January Evening Shift
MCQ (Single Correct Answer)
+4
-1
Change Language

The molar solubility(s) of zirconium phosphate with molecular formula $\left(\mathrm{Zr}^{4+}\right)_3\left(\mathrm{PO}_4^{3-}\right)_4$ is given by relation :

A
$\left(\frac{\mathrm{K}_{\mathrm{sp}}}{5348}\right)^{\frac{1}{6}}$
B
$\left(\frac{\mathrm{K}_{\mathrm{sp}}}{8435}\right)^{\frac{1}{7}}$
C
$\left(\frac{K_{s p}}{6912}\right)^{\frac{1}{7}}$
D
$\left(\frac{\mathrm{K}_{\mathrm{sp}}}{9612}\right)^{\frac{1}{3}}$
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