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100 mL solution of an acid (molar mass 8...

100 mL solution of an acid (molar mass 82) containing 39 g acid per litre was completely neutralized by 95.0 mL of aqueous NaOH solution containing 40 g NaOH per litre. The basicity of acid is ……

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The correct Answer is:
2
`N = ("gm eq")/(V_(L))`, gmeq of acid `= (g)/(E) = (29)/(82//x), " "N_(a) = (39 x)/(82) xx (1)/(1) = (39 x)/(82)`
Similarly `N_(b) = (g)/(E) xx (1)/(V_(L)) = (40)/(40) xx (1)/(1) = 1N`
For complete neutralization :
`N_(a)V_(a) = N_(b)V_(b) , " "(39 x)/(82) xx 100 = 1 xx 95 rArr x = 1.99 ~~ 2`
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