In the problem number 21, the number of mole of N_(2)O_(4) in 100 g of the mixture is:

In the problem number 21, the number of mole of N_(2)O_(4) in 100 g of

1.	In the problem number 21, the number of mole of N_(2)O_(4) in 100 g of the mixture is:
Answer» `0.43` `0.86` `0.57` `0.2` SOLUTION :`{:(,N_(2)O_(4),hArr,2NO_(2)),("At equilibrium",1-alpha,,2ALPHA):}` Number of MOLES of `N_(2)O_(4)=1-alpha=1-0.2=0.8` Weight of `N_(2)O_(4)` in MIXTURE =moles of `N_(2)O_(4)xx Mw` of `N_(2)O_(4)=0.8xx92=73.6 g` Weight of `NO_(2)` in mixture = moles of `NO_(2) xx Mw` of `NO_(2)` `=0.4xx46=18.4 g` Total weight `=73.6+18.4=92.0 g` In `92 g` of mixture, number of moles of `N_(2)O_(4)=0.8` In `100 g` of mixture, number of moles of `N_(2)O_(4)` `=(0.8xx100)/92=0.86`