Easy MCQ +4 / -1 PYQ · JEE Mains 2023

A cubic solid is made up of two elements X and Y. Atoms of X are present on every alternate corner and one at the center of cube. Y is at $\frac{1}{3}^{\mathrm{rd}}$ of the total faces. The empirical formula of the compound is :

  1. A $\mathrm{{X_2}{Y_{1.5}}}$
  2. B $\mathrm{{X_{3}}{Y_2}}$ Correct answer
  3. C $\mathrm{X{Y_{2.5}}}$
  4. D $\mathrm{{X_{2.5}}Y}$

Solution

$\begin{aligned} & \text { Number of } X \text { particles }=4 \times \frac{1}{8}+1=1.5 \\\\ & \text { Number of } Y \text { particles }=6 \times \frac{1}{3} \times \frac{1}{2}=1 \\\\ & \therefore \text { Empirical formula }=X_{1.5} Y_1=X_3 Y_2\end{aligned}$

About this question

Subject: Chemistry · Chapter: States of Matter · Topic: Gas Laws

This question is part of PrepWiser's free JEE Main question bank. 142 more solved questions on States of Matter are available — start with the harder ones if your accuracy is >70%.

Drill 25 more like these. Every day. Free.

PrepWiser turns these solved questions into a daily practice loop. Chapter-wise drills, full mocks, AI doubt chat. No auto-renew.

Start free →