Question

Ratio of active masses of 22g CO2,3g H2 and 7g
N2 in a gaseous mixture :-
(1) 22:37
(2) 0.5: 3:7
131.3.1
(4)1:3:0.5​

Answer 1

active mass is simply concentration, mostly molarity is assumed questions of chemical and ionic equilibrium.

22/44:3/2:7/28

=2:6:1

=1:3:0.5

Answer 2

Answer: (4)1:3:0.5​

Explanation:

Active mass is defined as the concentration of a reacting substance expressed usually in moles per liter.

Molarity of a solution is defined as the number of moles of solute dissolved per Liter of the solution.

$Molarity=\frac{n}{V_s}$

where,

n= moles of solute

$V_s$ = volume of solution  in Liters

a) 22 g of $CO_2$

$Moles=\frac{\text{Given mass}}{\text{Molar mass}}=\frac{22g}{44g/mol}=0.5mole$  

$Molarity=\frac{0.5}{V_s}$

b) 3 g of $H_2$

$Moles=\frac{\text{Given mass}}{\text{Molar mass}}=\frac{3g}{2g/mol}=1.5mole$  

$Molarity=\frac{1.5}{V_s}$

c) 7 g of $N_2$

$Moles=\frac{\text{Given mass}}{\text{Molar mass}}=\frac{7g}{28g/mol}=0.25mole$  

$Molarity=\frac{0.25}{V_s}$

Thus ratio of active masses of 22g $CO_2$ ,3g $H_2$ and 7g   $N_2$ in a gaseous mixture =$\frac{0.5}{V_s}$: $\frac{1.5}{V_s}$  : $\frac{0.25}{V_s}$  = 1: 3: 0.5

Thus the ratio is 1: 3: 0.5