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JEE Main 28-Jan-2025 Paper - Shift 1 — JEE STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 ScienceJEEJEE Main 28-Jan-2025 Paper - Shift 14 Marks
Question
Quantitative analysis of an organic compound (X) shows following % composition.
C : 14.5 %
$\mathrm{Cl}: 64.46 %$
H: 1.8 %
(Empirical formula mass of the compound $(\mathrm{X})$ is
\______________ $\times 10^{-1}$
$\overline{\text { (Given }}$ molar mass in $\mathrm{g} \mathrm{mol}^{-1}$ of $\mathrm{C}: 12, \mathrm{H}: 1$, $\mathrm{O}: 16, \mathrm{Cl}: 35.5$ )
✓

Answer

1655
Sol. $\quad \mathrm{C}: \mathrm{Cl}: \mathrm{H}: \mathrm{O}$
$\begin{array}{lllll}\text { %mass } & 14.5 & 64.46 & 1.8 & 19.24\end{array}$
Molar ratio $\frac{14.5}{12} \quad \frac{64.46}{35.5} \quad \frac{1.8}{1} \quad \frac{19.24}{16}$
$
\begin{array}{llll}
1.2 & 1.8 & 1.8 & 1.2
\end{array}
$
$\begin{array}{lllll}\text { Minimum } & 2 & 3 & 3 & 2\end{array}$
integral ratio
Empiricial formula $=\mathrm{C}_{2} \mathrm{H}_{3} \mathrm{Cl}_{3} \mathrm{O}_{2}$
Mass $=165.5$
Mass $=1655 \times 10^{-1}$

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