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

126 g

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118.5 g

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102.4 g

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110 g

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Solution

The correct option is C 118.5 g
Molar mass of $K_{2} {S O}_{4} . {A l}_{2} {({S O}_{4})}_{3} .24 H_{2} O = 948.7 g {m o l}^{- 1}$
No. of electrons involved $= 8$
$∴$ Equivalent weight $= \frac{948.7}{8} = 118.5 g$ .

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K_{2} {S O}_{4} . {A l}_{2} {({S O}_{4})}_{3} .24 H_{2} O

{({N H}_{4})}_{2} {S O}_{4} . {A l}_{2} {({S O}_{4})}_{3} .24 H_{2} O

H_{2} {S O}_{4}

(a) H_{2} {S O}_{4} + N a O H \to N a H {S O}_{4} + H_{2} O

(b) H_{2} {S O}_{4} + 2 N a O H \to {N a}_{2} {S O}_{4} + 2 H_{2} O

2.5 \times 10^{- 2} g

K_{2} {S O}_{4}

K^{- 1} {m o l}^{- 1}

K_{2} {S O}_{4}

{m o l}^{- 1}

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