Question

How many mL of concentrated sulphuric acid of sp. gr. 1.84 containing 98% H2SO4 solution by weight is required to prepare 200 mL of 0.5 N solution?

Answer 1

Answer:

molecular weight of sulfuric acid = 98g

Specific Gravity  = 1.84

% of pure sulphuric acid/gm = 98%

number of replacable hydrogens = 2

Gram Equivalent Weight =molecular weight/replaceable hydrogens

98/2 = 49g

 

Required substance weight = eq. wt. x Normality x Volume (in L)

=49gm x0.5N X 0.2 Litres = 4.9 g  

Dilution Factor = Specific Gravity x concentration of pure acid.

=1.84 x 0.98 = 1.8gm  

Volume required = mass required/dilution factor  

4.9/1.8 = 2.72ml.

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