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JEE Advanced Chemistry Electrochemistry 2022 JEE Advanced 2022 (Paper 2)

JEE Advanced Chemistry Question (2022) — Solution

Question

Consider the strong electrolytes Z m X n ,   U m Y p  and V m X n . Limiting molar conductivity of U m Y p and V m X n are 250 and 440   S   cm 2   mol - 1 , respectively. The value of m + n + p is Given: Ion Z n + U p + V n + X m - Y m - λ 0 S   cm   2 mol - 1 50 . 0 25 . 0 100 . 0 80 . 0 100 . 0 λ 0  is the limiting molar conductivity of ions The plot of molar conductivity Λ of Z m X n vs c 1 2 is given below. If the numerical value has more than two decimal places, truncate/round-off the value to TWO decimal places.

Options

  1. A. A
  2. B. B
  3. C. C
  4. D. D

Answer

A. A

Step-by-step solution

λ m = λ m ° - A C For electrolyte Z m X n  and from given curve λ m Z m X n = λ m 0 Z m X n - A C - A = 336 - 339 0 . 04 - 0 . 01 = - 3 0 . 03 ⇒ A = 100 ∴ For λ m = 336   S   cm 2   mol - 1 ⇒ 336 = λ m 0 Z m X n - 100 × 0 . 04 λ m 0 = 336 + 4 = 340   S   cm 2   mol - 1 Z m X n ⟶ mZ n + + nX m - ∴    50 m + 80 n = 340 ⇒ 5 m + 8 n = 34                   . . . i U m Y p ⟶ mU p + + pY m - ∴ 25   m + 100 p = λ m 0 U m Y p = 250 ⇒ m + 4 p = 10           . . . ii V m X n ⟶ mV n + + nX m - ∴ 100 m + 80 n = 440 ⇒ 5 m + 4 n = 22             . . . iii From equation i and iii n = 3 m = 2 From equation ii p = 2 ∴    m + n + p = 2 + 3 + 2 = 7

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