Chemistry · 2026
JEE Main · 2 April 2026, Shift 1 · Q56
The solubility product constants of Ag_2 CrO_4 and AgBr are 32 x and 4 y respectively at 298 K. The value of ((molarity of Ag_2 CrO_4)/(molarity of…
The solubility product constants of $\displaystyle \mathrm{Ag}_2 \mathrm{CrO}_4$ and AgBr are $\displaystyle 32 x$ and $\displaystyle 4 y$ respectively at $\displaystyle 298$ K . The value of $\displaystyle \left(\frac{\text { molarity of } \mathrm{Ag}_2 \mathrm{CrO}_4}{\text { molarity of } \mathrm{AgBr}}\right)$ can be expressed as :
Official answer
From NTA’s final answer key for this paper.
(4)
$\displaystyle \frac{\sqrt[3]{x}}{\sqrt{y}}$
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JEE Main 2026 Chemistry question, with the answer from NTA’s final answer key. Where our answers come from.