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Chemistry
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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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
Some Basic Concepts of Chemistry
A
$0.2\, N$
25%
B
$0.3 \,N$
0%
C
$0.6 \,N$
75%
D
$1.8 \,N$
0%
Solution:
$FaCl _3 \equiv Fe ( OH )_3$
$M \times 100 \equiv \frac{2.14}{107} \times 1000$
$M_{ FeCl _3}=0.2$
$N_{ FeCl _3}=0.2 \times 3($ valency factor $=3)$
$=0.6 \,N$