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Question

How many moles of Al3+ are present in 300mL of 0.2 N Al2(SO4)3?

A
2×102
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B
1×103
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C
1×102
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D
1×104
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Solution

The correct option is B 2×102
nfactor of Al=3 (Valency)
Numberofmoles=Normality×Volumenfactor=0.2×3003=20 millimoles=2×102 moles

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