1.

The mass of 70% `H_(2)SO_(4)` required for neutralization of one mole of NaOH is:A. 49 gB. 98 gC. 70 gD. `34*3 g`

Answer» Correct Answer - 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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