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Complex 'A' has composition of ${CrCl}_{3} H_{12} O_{6}$ . Complex on treatment with con $H_{2} S O_{4}$ . loses $13.5 %$ of its original mass.The correct formula of 'A' is (given atomic mass of $C r = 52$ , $C l = 35.5)$

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$[Cr {(H_{2} O)}_{6}] {Cl}_{3}$

$[Cr {(H_{2} O)}_{4} {Cl}_{2}] Cl .2 H_{2} O$

$[Cr {(H_{2} O)}_{3} {Cl}_{3}] .3 H_{2} O$

$[Cr {(H_{2} O)}_{5} Cl] {Cl}_{2} . H_{2} O$

answer is B.

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Detailed Solution

Molar mass of ${CrCl}_{3} H_{12} O_{6}$ is 266.5
It loses 13.5% mass due to water which is 36g.
Therefore, the correct formula is $[Cr {(H_{2} O)}_{4} {Cl}_{2}] Cl .2 H_{2} O$

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