(i)[Ag+] required to ppt AgCl(s)
Ksp=IP=[Ag+][Cl−]=1.7×10−10
[Ag+]=1.7×10−9
(ii)[Ag+] required to ppt Ag2CrO4(s)
Ksp=IP=[Ag+]2[CrO4−2]=1.9×10−12
[Ag+]=4.3×10−5
[Ag+] required to ppt AgCl is low so AgCl will ppt 1st.
A solution is 0.1M in Cl^- and 0.001M in CrO_4^2-. Solid AgNO_3 is gradually added to it Assuming that the addition does not change in volume and K_…
JEE Main 2021 — Chemistry Physical Chemistry
2021mcqhard
A solution is 0.1M in Cl− and 0.001M in CrO42−.
Solid AgNO3 is gradually added to it Assuming that the addition does not change in volume and Ksp(AgCl)=1.7×10−10M2 and
Ksp(Ag2CrO4)=1.9×10−12M3.
Select correct statement from the following:
Official previous-year question
Held on 20 Jul 2021 · Verified 6 Jul 2026.
Options
- A
AgCl precipitates first because its Ksp is high.
- B
Ag2CrO4 precipitates first as its Ksp is low.
- C
Ag2CrO4 precipitates first because the amount of Ag+ needed is low.
- D
AgCl will precipitate first as the amount of Ag+ needed to precipitate is low.
Did you get this right?
Sign in to track your attempts and accuracy.
Your note
Sign in to keep a private note on this question. Nothing you write is ever public.
JEE Main Physical Chemistry in other years
More JEE Main Physical Chemistry questions
What is the ratio of wave number of first line (lowest energy line) of Balmer series of H atomic spectrum to first line of its Brackett series?
2026medium
  Which of the following point in Figure 2 most accurately represents the nodal surface as shown in Figure 1?
2026medium
At 298 K, the mole percentage of $\mathrm{N}_{2}(\mathrm{~g})$ in air is $80 \%$. Water is in equilibrium with air at a pressure of 10 atm. What is the mole fraction of $\mathrm{N}_{2}(\mathrm{~g})$ in water at 298 K ? ($\mathrm{K}_{\mathrm{H}}$ for $\mathrm{N}_{2}$ is $6.5 \times 10^{7} \mathrm{~mm} \mathrm{Hg}$)
2026medium
Electricity is passed through an acidic solution of $\mathrm{Cu}^{2+}$ till all the $\mathrm{Cu}^{2+}$ was exhausted, leading to the deposition of 300 mg of Cu metal. However, a current of 600 mA was continued to pass through the same solution for another 28 minutes by keeping the total volume of the solution fixed at 200 mL. The total volume of oxygen evolved at STP during the entire process is $\_\_\_\_$ mL. (Nearest integer) [Given: $\mathrm{Cu}^{2+}(\mathrm{aq})+2 \mathrm{e}^{-} \rightarrow \mathrm{Cu}(\mathrm{s}) \mathrm{E}_{\text {red }}^{\mathrm{o}}=+0.34 \mathrm{~V}$ $\mathrm{O}_{2}(\mathrm{~g})+4 \mathrm{H}^{+}+4 \mathrm{e}^{-} \rightarrow 2 \mathrm{H}_{2} \mathrm{O} \mathrm{E}_{\text {red }}^{\mathrm{o}}=+1.23 \mathrm{~V}$ Molar mass of $\mathrm{Cu}=63.54 \mathrm{~g} \mathrm{~mol}^{-1}$ Molar mass of $\mathrm{O}_{2}=32 \mathrm{~g} \mathrm{~mol}^{-1}$ Faraday Constant $=96500 \mathrm{C} \mathrm{mol}^{-1}$ Molar volume at $\mathrm{STP}=22.4 \mathrm{~L}$]
2026hard
Arrange the following resultant mixtures in increasing order of their pH values A. $10$ mL $0.2$ M Ca(OH)$_2$ + $25$ mL $0.1$ M HCl B. $10$ mL $0.01$ M H$_2$SO$_4$ + $10$ mL $0.01$ M Ca(OH)$_2$ C. $10$ mL $0.1$ M H$_2$SO$_4$ + $10$ mL $0.1$ M KOH Choose the correct answer from the options given below:
2026medium
Other Chemistry topics for JEE Main
Keep practicing Physical Chemistry
Work through every JEE Main Physical Chemistry PYQ, year by year.