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The mass of 80% pure H(2)SO(4) required ...

The mass of `80%` pure `H_(2)SO_(4)` required to completely neutralise 60g of `NaOH` is 

A

`92g`

B

`58.8g`

C

`73.5g`

D

`98g`

Text Solution

Verified by Experts

The correct Answer is:
A
`2NaOH+H_(2)SO_(4)rarrNa_(2)SO_(4)+H_(2)O`
`2rarr1`
`1.5rarr?`
`H_(2)SO_(4)` = 0.75 moles = 73.5 gm
80% W = 73.5 implies W = 92 gm
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