Question

Ratio of active masses of 22g carbon dioxide, 3g Hydrogen and 7g Nitrogen in a gaseous mixture ?

Answer 1

1:
 Number of moles of CO2 = given weight/molecular weight
= 22/44
 = ½
Number of moles per V liters of volume = ½V
 2:
Number of moles of H2 = given weight/molecular weight
= 3/2
Number of moles per V liters of volume = 3/2V
3:
Number of moles of N2 = given weight/molecular weight
= 7/28
 = ¼
Number of moles per V liters of volume = ¼V

Total number of moles = nCO2 + nH2 + nN2
= ½V + 3/2V +¼V
= 9/4V

Ratio of number of moles in comparison with total number of moles:
 = (½V)/(9/4V) : (3/2V)/(9/4V) : (¼V)/(9/4V)
= 2/9 : 6/9 : 1/9

Ratio of active masses of gases = 2 : 6 : 1

Answer 2

Hello there,

Active mass is defined in terms of molarity of the substance(other than solid because the active mass of solid is always 1).

Let volume of all the given gases is V.

Therefore no. of molarity of all the gases is in general

(No. of moles/V)

Therefore molarity of Carbon dioxide is (22/44V)=0.5/V.

So molarity of Hydrogen is (3/2V)=1.5/V,

and molarity of Nitrogen is (7/28V)=0.25/V,

Therefore the ratio of active masses are

0.5:1.5:0.25

=1:3:0.5

Hope it help you!