Chemistry miscellaneousHard

Question

The mass of 70% pure H2SO4 required for neutralisation of 500 ml of 2M of NaOH is :-

Options

A.49 g
B.98 g
C.70 g
D.24.5 g

Solution

Meq. of H2SO4 = meq. of NAOH
= N × V ⇒ W = 49 g
But H2SO4 is 70% pure therefore weight
= 49 × = 70 g

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