The value of $x$ in $K A l (S O_{4})_{x} \cdot 12 H_{2} O$ is:

1

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2

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3

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4

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Solution

The correct option is A $2$
The value of $x$ in $K A 1 (S O_{4})_{x} \cdot 12 H_{2} O$ is $2$ .
$K$ and $A l$ have charges $+ 1$ and $+ 3$ respectively. Total charge on cations is $1 + 3 = 4$ .
Each sulphate ion has charge of $- 2$ . To balance $+ 4$ charge on cations with $- 2$ charge on sulphate ions, two sulphate ions are required.
Hence, the option $(B)$ is the correct answer.

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K A l (S O_{4})_{x}, 12 H_{2} O

K A l (S O_{4})_{x}, 12 H_{2} O

a

b

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(1 + x + 2 x^{2} + 3 x^{3} + 4 x^{4})^{4}

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(x - 1) - \frac{1}{2} (x - 1)^{2} + \frac{1}{3} (x - 1)^{3} - \frac{1}{4} (x - 1)^{4} + \dots .

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⎡ ⎢ ⎣ \begin{matrix} 3 & - 1 + x & 2 3 & - 1 & x + 2 x + 3 & - 1 & 2 \end{matrix} ⎤ ⎥ ⎦

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