Question bank/JEE Main/Chemistry

1 g of complex was passed through a cation
exchanger to produce HCl. The acid liberated was diluted to 1 liter.
What will be the molarity of acid solution [Molecular weight of complex
= 266.5] ?

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1 mole of complex will give two moles of Cl- ion, i.e. 2 mole
HCl
$\lbrack Cr{(H_{2}O)}_{5}Cl\rbrack{Cl}_{2} \cdot H_{2}O \rightleftharpoons {\lbrack Cr{(H_{2}O)}_{5}Cl\rbrack}^{+ 2} + 2{Cl}^{-}$
1g
mole of HCl$= \frac{2 \times 1}{266.5}$ (for HCl, V.f. = 1
[IMAGE] N = M)
$\therefore N = \frac{2/266.5}{1} = 0.0075$

Source: Grass-G / CH-15 Chemistry Paper 11 Nov..docx