JEE Mains · Chemistry · STD 12 - 5. Co-ordination chemistry
Complex \(A\) has a composition of \(H _{12} O _{6} Cl _{3} Cr\). If the complex on treatment with conc. \(H _{2} SO _{4}\) loses \(13.5 \%\) of its original mass, the correct molecular formula of \(A\) is : [Given : atomic mass of \(Cr =52\) \(amu\) and \(Cl =35\, amu ]\)
- A \(\left[ Cr \left( H _{2} O \right)_{5} Cl \right] Cl _{2} \cdot H _{2} O\)
- B \(\left[ Cr \left( H _{2} O \right)_{3} Cl _{3}\right] \cdot 3 H _{2} O\)
- C \(\left[ Cr \left( H _{2} O \right)_{4} Cl _{2}\right] Cl \cdot 2 H _{2} O\)
- D \(\left[ Cr \left( H _{2} O \right)_{6}\right] Cl _{3}\)
Answer & Solution
Correct Answer
(C) \(\left[ Cr \left( H _{2} O \right)_{4} Cl _{2}\right] Cl \cdot 2 H _{2} O\)
Step-by-step Solution
Detailed explanation
\(\%\) 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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