Home Maths JEE - Advanced Previous Year Paper JEE-Advanced-2022-Paper-2 CHEMISTRY SECTION 1 (Maximum Marks: 24) • Th…
Maths JEE - Advanced Previous Year Paper JEE-Advanced-2022-Paper-2 Numeric Response
Published on: August 13, 2026

CHEMISTRY

SECTION 1 (Maximum Marks: 24)

• This section contains EIGHT (08) questions.

• The answer to each question is a SINGLE DIGIT INTEGER ranging from 0 TO 9, BOTH INCLUSIVE.

• For each question, enter the correct integer corresponding to the answer using the using the mouse and the onscreen virtual numeric keypad in the place designated to enter the answer.

• Answer to each question will be evaluated according to the following marking scheme:

Full Marks : +3 If ONLY the correct integer is entered;

Zero Marks : 0 If the question is unanswered;

Negative Marks : -1 In all other cases.

Concentration of H 2 SO 4 and Na 2 SO 4 in a solution is 1M and l.8x10 -2 M, respectively. Molar solubility of PbSO 4 in the same solution is Xx10 -Y M (expressed in scientific notation). The value of Y is_________. [Given : Solubility product of PbSO 4 (K sp ) = 1.6x10 -8 . For H 2 SO 4 , K a1 is very large and K a2 =1.2x10 -2 ]

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Text Solution

Verified by Experts
The correct answer is:
6

(6)

Now,

a=0.006

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