Question

0.0012 mole of $CrC{l}_3\cdot 6H_{2}O$ was passed through a cation exchange resin and acid coming out of it required 28.5 mL of 0.125 M $NaOH$. Hence, the formula of the complex is:

• A. $\left[Cr\right(H_{2}O{)}_{5}Cl]C{l}_2\cdot H_{2}O$
• B. $\left[Cr\right(H_{2}O{)}_{4}C{l}_2]\cdot 2H_{2}O$
• C. $\left[Cr\right(H_{2}O{)}_6]C{l}_3$
• D. $\left[Cr\right(H_{2}O{)}_{3}C{l}_3]\cdot 3H_{2}O$

Answer 1

The correct option is C $\left[Cr\right(H_{2}O{)}_6]C{l}_3$

The number of moles of NaOH are $\frac{28.5}{1000}\times 0.125=0.0036$.

The ratio of the number of moles of $NaOH$ to the number of moles of $CrC{l}_3\cdot 6H_{2}O$ is $\frac{0.0036}{0.0012}=3:1$.

Thus, 1 mole of $CrC{l}_3\cdot 6H_{2}O$ reacts with 3 moles of $NaOH$ i.e. 1 mole of complex has 3 $C{l}^{-}$ ions to form 3 $NaCl$ with 3 moles of $NaOH$.
Hence, the formula of $CrC{l}_3\cdot 6H_{2}O$ is

$\left[Cr\right(H_{2}O{)}_6]C{l}_3$.