Question:
Complex A has a composition of $\mathrm{H}_{12} \mathrm{O}_{6} \mathrm{Cl}_{3} \mathrm{Cr}$. If the complex on treatment with conc. $\mathrm{H}_{2} \mathrm{SO}_{4}$ loses $13.5 \%$ of its original mass, the correct molecular formula of $\mathrm{A}$ is :
[Given : atomic mass of $\mathrm{Cr}=52 \mathrm{amu}$ and $\mathrm{Cl}=35 \mathrm{amul}$
Correct Option: , 3
Solution:
$\%$ mass of water
$=\frac{x \times 18}{(12+6 \times 16+35 \times 3+52)} \times 100=13.5$
$\Rightarrow \quad x=\frac{265 \times 13.5}{18 \times 100} \simeq 2$
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