The normility of a solution that results from mixing `4g` of `NaOH, 500 mL` of `1 M HCl`, and `10.0mL` of `H_(2)O_(4)` (specific gravity 1.149% `H_(2) SO_(4)` by weight) is
The total volume of solution was made to `1 L` with water)

Options

A. 0.51
B. 0.71
C. 1.02
D. 0.45

Answer

A

Explanation

`4 g "of" NaOH = 4//40 = 0.1 "mol" = 100 "mmol" -= 100 mEq`
`HCl = 500 xx 1 = 500 mEq`
`N "of" H_(2)SO_(4) = (W_(2) xx 1000)/(Ew_(2) xx V_(sol) ("in" mL))`
or
`= (% "byWeight" xx 10 xx d)/(Ew_(2))`
`= (49 xx 10 xx 1.1)/(49) = 11 N`
`mEq "of" H_(2)SO_(4) = 11 N xx 10.00 mL = 110 mEq`
Total acid `= 110 + 500 = 610 mEq`
`NaOH = 100 mEq`
Acid left `= 610 - 100 = 510 mEq`
Total volume `= 1000 mL`
Normality of solution `= (mEq)/(mL) = (510)/(1000) = 0.51 N`