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1.176 g of sulphuric acid is present in ...

1.176 g of sulphuric acid is present in 200 m of its solution. Calculate the normality of the solution.
(Equivalent mass of sulphuric acid = 49)

Text Solution

Verified by Experts

1.176 g of sulphuric acid is present in 200 mL.
1000 mL contains `(1.176xx1000)/(200)=5.88` g of sulphuric acid.
Therefore, `N=W/(E)=5.88/(49)=0.12`
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