JEE Advanced 2022 — Paper 2 — CHEM-II solution

JEE Advanced2022Paper 2CHEM-II

To check the principle of multiple proportions, a series of pure binary compounds (P $_{m}$ Q $_{n}$ ) were analyzed and their composition is tabulated below. The correct option(s) is(are)

Compound	Weight % of P	Weight % of Q
1	50	50
2	44.4	55.6
3	40	60

A
If empirical formula of compound 3 is P $_{3}$ Q $_{4}$ , then the empirical formula of compound 2 is P $_{3}$ Q $_{5}$ .
B
If empirical formula of compound 3 is P $_{3}$ Q $_{2}$ and atomic weight of element P is 20, then the atomic weight of Q is 45.
C
If empirical formula of compound 2 is PQ, then the empirical formula of the compound 1 is P $_{5}$ Q $_{4}$ .
D
If atomic weight of P and Q are 70 and 35, respectively, then the empirical formula of compound 1 is P $_{2}$ Q.

Solution

Let $M_{P}$ , $M_{Q}$ be the atomic weights of P and Q.

(A) If compound 3 is P $_{3}$ Q $_{4}$ with 40% P, then $\frac{3 M _{P}}{3 M _{P} + 4 M _{Q}} = 0.40$ , giving $9 M_{P} = 8 M_{Q}$ . For P $_{3}$ Q $_{5}$ , the predicted % of P is $\frac{3 M _{P}}{3 M _{P} + 5 M _{Q}} = \frac{8 M _{Q} /3}{8 M _{Q} /3 + 5 M _{Q}} \approx 34.8%$ , not 44.4%. Incorrect.

(B) If compound 3 is P $_{3}$ Q $_{2}$ with 40% P, then $\frac{3 M _{P}}{3 M _{P} + 2 M _{Q}} = 0.40$ , giving $9 M_{P} = 4 M_{Q}$ . If $M_{P} = 20$ , then $M_{Q} = 180/4 = 45$ . Correct.

(C) If compound 2 is PQ with 44.4% P: $\frac{M _{P}}{M _{P} + M _{Q}} = \frac{4}{9}$ , so $5 M_{P} = 4 M_{Q}$ . Predicted % of P in P $_{5}$ Q $_{4}$ : $\frac{5 M _{P}}{5 M _{P} + 4 M _{Q}} = \frac{5 M _{P}}{5 M _{P} + 5 M _{P}} = 50%$ . Matches compound 1. Correct.

(D) If P $_{2}$ Q for compound 1 with 50% P: $\frac{2 M _{P}}{2 M _{P} + M _{Q}} = 0.5$ , so $M_{Q} = 2 M_{P}$ . With $M_{P} = 70$ , $M_{Q}$ should be 140, not 35. Incorrect.

$∴$ (B), (C).

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