To `1 L` of `1.0 M` impure `H_(2)SO_(4)` sample, `1.0 M NaOH` solution was added and a plot was obtained as follows:
The % purity of `H_(2) SO_(4)` and the slope of curve, respectively, are:

Options

A. `75%, - 1//2`
B. `75%, - 1`
C. `50%, -1//3`
D. `50%, -1//2`

Answer

B

Explanation

Moles of `NaOH = 1.0 M xx 1.5 L = 1.5 "mol"`
Moles of `H_(2)SO_(4) = 1.0 M xx 1 L = 1.0 "mol"`
`underset(2 "mol")(2NaOH) + underet(2 "mol")(H_(2) SO_(4)) rarr Na_(2)SO_(4) + 2H_(2) O`
Moles of `H_(2)SO_(4)` reacted with `NaOH`
`(1)/(2) "mol of" NaOH`
`= (1)/(2) xx 1.5 = 0.75 "mol" H_(2) SO_(4)`
% purity of `H_(2) SO_(4) = (0.75 "mol" xx 100)/(1.0 "mol") = 75%`
ii. For slope: `(x)/(a) + (y)/(b) = 1`
Slope `((y)/(x)) = (-b)/(a) = - (1.5)/(1.5) = - 1`