JEE PYQ: Coordination Compounds Question 20
Question 20 - 2020 (03 Sep Shift 2)
Complex A has a composition of $\text{H}_{12}\text{O}_6\text{Cl}_3\text{Cr}$. If the complex on treatment with conc. $\text{H}_2\text{SO}_4$ 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) $[\text{Cr(H}_2\text{O})_6]\text{Cl}_3$ (b) $[\text{Cr(H}_2\text{O})_3\text{Cl}_3] \cdot 3\text{H}_2\text{O}$ (c) $[\text{Cr(H}_2\text{O})_5\text{Cl}]\text{Cl}_2 \cdot \text{H}_2\text{O}$ (d) $[\text{Cr(H}_2\text{O})_4\text{Cl}_2]\text{Cl} \cdot 2\text{H}_2\text{O}$
Show Answer
Answer: (c)
Solution
Molar mass = 265. Loss = 13.5% of 265 = 35.8 ≈ 36 = 2 H$_2$O molecules. $[\text{Cr(H}_2\text{O})_4\text{Cl}_2]\text{Cl} \cdot 2\text{H}_2\text{O}$ loses 2 water = 36/265 = 13.6%. Answer key indicates option (3).
BETA
