NEET UG 2024 — Chemistry Physical Chemistry

2024mcqeasy

A compound $X$ contains $32 \%$ of $A, 20 \%$ of $B$ and remaining percentage of $C$ . Then, the empirical formula of X is : (Given atomic masses of $A=64 ; B=40 ; C=32 u$ )

Official previous-year question

Held on 30 Apr 2024 · Verified 9 Jul 2026.

Options

A
$\mathrm{ABC}_3$
B
$\mathrm{AB}_2 \mathrm{C}_2$
C
$\mathrm{ABC}_4$
D
$\mathrm{A}_2 \mathrm{BC}_2$

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Solution

$\begin{array}{|c|c|c|c|c|}\hline \text{Element} & \begin{array}{c} \text{Mass} \\ \text{percentage} \%\end{array} & \begin{array}{c} \text{No. of} \\ \text{moles} \end{array} & \begin{array}{c} \text{No. of moles} / \\ \text{Smallest number} \end{array} & \begin{array}{c} \text{Simplest whole} \\ \text{number} \end{array} \\\hline A & 32 \% & \frac{32}{64}=\frac{1}{2} & \frac{1}{2} \times 2 & =1 \\\hline B & 20 \% & \frac{20}{40}=\frac{1}{2} & \frac{1}{2} \times 2 & =1 \\\hline C & 48 \% & \frac{48}{32}=\frac{3}{2} & \frac{3}{2} \times 2 & =3 \\\hline\end{array}$ So, empirical formula of $X=\begin{array}{clll}A & : & B & \text { C } \\ 1 & : & 1 & :\end{array}$ $\therefore$ The correct empirical formula of compound X is $\mathrm{ABC}_3$

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NEET UG 2024 — Chemistry Physical Chemistry

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NEET UG 2024 — Chemistry Physical Chemistry

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