What is the strength in gram per litre of a solution of `H_(2)SO_(4), 12 mL` of which neutralised by `15 mL` of `N//10` `NaOH` solution?

Answer

A, B

Explanation

mEq of `H_(2) SO_(4) =` mEq of `NaOH`
`N_(1) V_(1) = N_(2) V_(2)`
`N_(1) xx 12 (1)/(10) xx 15`
`:. N_(2) = 0.125`
Strength `= N xx Ew = 0.125 xx 49 = 6.125 g L^(-1)`
`[{:(Mw of H_(2) SO_(4) = 1 xx 2 + 32 + 16 xx 4 = 98 g),(Ew = (Mw)/(2) = (98)/(2) = 49 g):}]`