The half life for the reaction `N_(2)O_(5) hArr 2NO_(2)+1/2 O_(2)` in `24 hr` at `30^(@)C`. Starting with `10 g` of `N_(2)O_(5)` how many grams of `N_(2)O_(5)` will remain after a period of `96` hours ?

Options

A. `1.25` g
B. `0.63` g
C. `1.7` g
D. `0.5` g

Answer

B

Explanation

`k = (0.693)/(24)hr^(-1) = (2.303)/(96)"log"(10)/(a-x)`
or `"log"(10)/(a-x) = 1.2036`
or `1 - log(a-x) = 1.2036`
or `log(a-x) = -0.2036 = 1.7964`
or `(a-x) = "anitlog" 1.7964 = 0.6258g`