Question

The `P^(H)` of a sample of `H_(2)SO_(4)` is `1.3979`. The percentage of the solution is `73.5% (w//w)`, the density of the solution is
A. `2.66 xx 10^(-3) g//c c`
B. `5.32 xx 10^(-3) g// c c`
C. `1.33 xx 10^(-3) g// c c`
D. `0.01 g//c c`

Answer 1

Correct Answer - A
`[H^(+)]` = Normality of `H_(2)SO_(4)`
`= 10^(-2) xx` anti log of `0.6021 = 4 xx 10^(-2)`
But noramality ` = (% xx d xx 10)/("Ew.Wt")`
`rArr 4 xx 10^(-2) = (73.5 x d xx 10)/(49)`
`:. D = 2.66 xx 10^(-3) gm//c c`