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The mass of 70% H(2)SO(4) required for n...

The mass of 70% `H_(2)SO_(4)` required for neutralization of one mole of NaOH is:

A

49 g

B

98 g

C

70 g

D

`34*3 g`

Text Solution

Verified by Experts

The correct Answer is:
C
`2NaOH + H_(2)SO_(4) to Na_(2)SO_(4)+ H_(2)O`
Pure `H_(2)SO_(4)` required for 2 mole of NaOH
`=(49xx100)/(70) = 70g`
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