Solve this following


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 $\mathrm{Cl}=35 \mathrm{amul}$


  1. $\left[\mathrm{Cr}\left(\mathrm{H}_{2} \mathrm{O}\right)_{5} \mathrm{Cl}\right] \mathrm{Cl}_{2} \cdot \mathrm{H}_{2} \mathrm{O}$

  2. $\left[\mathrm{Cr}\left(\mathrm{H}_{2} \mathrm{O}\right)_{3} \mathrm{Cl}_{3}\right] \cdot 3 \mathrm{H}_{2} \mathrm{O}$

  3. $\left[\mathrm{Cr}\left(\mathrm{H}_{2} \mathrm{O}\right)_{4} \mathrm{Cl}_{2}\right] \mathrm{Cl} \cdot 2 \mathrm{H}_{2} \mathrm{O}$

  4. $\left[\mathrm{Cr}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right] \mathrm{Cl}_{3}$

Correct Option: , 3


$\%$ 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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