Solve this: 36 At the equilibrium of the reaction, N2O4 ⇌ 2NO2(g), the observed molar mass of - Chemistry - Electrochemistry

Question

Solve this:

36     A t   t h e   e q u i l i b r i u m
  o f   t h e   r e a c t i o n ,   N 2 O 4
  ⇌   2 N O 2 ( g ) ,   t h e   o b s e r
v e d   m o l a r   m a s s   o f   N 2 O 4
    i s   77 70   g   m o l - 1   T h
e   p e r c e n t a g e   d i s s o c i a t i o n  
o f   N 2 O 4     i s
(1) 28 4 (2) 46 7 (3) 22 4 (4) 18 4

Megha Sharma · Accepted Answer

Dear student,
Van't hoff factor = Observed molar mass / normal molar mass
= 92 / 77 7
= 1 184
Van't hoff factorâ€‹ = 1 +
αâ€‹ (n-1)
1 184 = 1 + αâ€‹ ( 2 - 1)
1 184 - 1 = αâ€‹
αâ€‹ = 0 184
% dissociation = αâ€‹ * 100 = 0 184 * 100 =
18 4 %
So the answer is (4) i e 18 4%
Regards

Solve this: 36 . A t t h e e q u i l i b r i u m o f t h e r e a c t i o n , N 2 O 4 ⇌ 2 N O 2 ( g ) , t h e o b s e r v e d m o l a r m a s s o f N 2 O 4 i s 77 . 70 g m o l - 1 . T h e p e r c e n t a g e d i s s o c i a t i o n o f N 2 O 4 i s (1) 28.4 (2) 46.7 (3) 22.4 (4) 18.4

Solve this:

$36 . A t t h e e q u i l i b r i u m o f t h e r e a c t i o n, N_{2} O_{4} ⇌ 2 N O_{2} (g), t h e o b s e r v e d m o l a r m a s s o f N_{2} O_{4} i s 77.70 g m o l^{- 1} . T h e p e r c e n t a g e d i s s o c i a t i o n o f N_{2} O_{4} i s$
(1) 28.4 (2) 46.7 (3) 22.4 (4) 18.4