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Q. An excess of $NaOH$ was added to $100 \,mL$ of a $FeCl _3$ solution which gives $2.14$ of $Fe ( OH )_3$. Calculate the normality of $FeCl _3$ solution

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Solution:

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$FaCl _3 \equiv Fe ( OH )_3$
$M \times 100 \equiv \frac{2.14}{107} \times 1000$
$M_{ FeCl _3}=0.2$
$N_{ FeCl _3}=0.2 \times 3($ valency factor $=3)$
$=0.6 \,N$