Question

0.63 gram oxalic acid (equivalent weight = 63) is dissolved in 250 ml
of solution. Find out the normality of solution.​

Answer 1

Given,

m=0.63g

Eq. wt. =63

Volume =250/1000 L

=1/4L

Therefore,

Normality (N)= m/eq. wt.×vol.

=63/0.63×(1/4)

=0.04 N

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Answer 2

Given Info : 0.63 g oxalic acid ( equivalent weight is 63 ) is dissolved in 250 ml of solution.

To find : the normality of solution is ...

solution : volume of solution = 250 ml = 250/1000 l [ ∵ 1000 ml = 1 litre ]

= 0.25 l

no of equivalents = given weight/equivalent weight

given weight of oxalic acid = 0.63 g

equivalent weight of oxalic acid = 63 g

∴ no of equivalents of oxalic acid = 0.63/63 = 0.01

now normality = no of equivalents/volume of solution in L

= 0.01/0.25

= 0.04 N

Therefore the normality of solution is 0.04 N.

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