Question

Normality of 0.98% H2SO4 solution is

Answer 1

0.98%w/v means 0.98 g H2SO4 in 100 ml of solution

No. of gram equivalents= (mass/ Equivalent mass)=0.98/49 = 0.02 moles

                                     { Eq.mass=Mol.wt/valency=98/2=49}

Normality=No. of Equivalents*(1000/ v in ml)

               0.02*1000/100=0.2 N

Answer 2

Answer:

Normality of 0.98% H2SO4 solution is 0.2 N.

Explanation:

Given: 0.98% H2SO4 solution

Find:Normality

step by step explanation

Normality=1000×W/EV

where W is the mass of sulphuric acid, E is the equivalent weight and V is the volume of solution.

We know, Sulphuric acid is a dibasic acid, hence, its Equivalent weight: = Molecular weight/ 2 = 49.

So, Normality = (0.98 x 1000) / (49 x 100)

= 0.2 N

Hence Normality of 0.98% H2SO4 solution is 0.2 N.

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