SATHEE: Co-Ordination Compounds Question 48

Co-Ordination Compounds Question 48

Question: The formula of alum is

[Pb. CET 2002]

Options:

A) $K_{2} S O_{4} . A l_{2} {(S O_{4})}_{3} .24 H_{2} O$

B) $K_{4} [F e {(C N)}_{6}]$

C) $K_{2} S O_{4} . A l_{2} {(S O_{4})}_{3} .6 H_{2} O$

D) $N a_{2} C O_{3} .10 H_{2} O$

Show Answer

Answer:

Correct Answer: A

Solution:

General formula for alum is $M_{2} S O_{4} . R_{2} {(S O_{4})}_{3} .24 H_{2} O$

$M =$ mono valent cation $(K^{+},, N a^{+} \dots .)$

$R =$ Trivalent cation $(A l^{+ 3},, F e^{+ 3})$ Hence, $K_{2} S O_{4} A l_{2} {(S O_{4})}_{2} .24 H_{2} O$ represent an alum.

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