The vant Hoff factor of BaCl2 at 0.01 M concentration is 1.98 . The percentage of dissociation of BaCl2 at this concentration is:

Question

The vant Hoff factor of BaCl2 at 0.01 M concentration is 1.98 . The percentage of dissociation of BaCl2 at this concentration is: (A) 49 (B) 69 (C) 89 (D) 98

Tardigrade Admin · Accepted Answer

BaCl2 leftharpoons Ba2++2Cl- Initial 0.01M At equilibrium (0.01-x)MxM2xM i=((0.01-x)+x+2x/0.01) =(0.01+2x/0.01)=1.98 x=0.0049 % α=(x/0.01)× 100=(0.0049× 100/0.01)=49 %

Q. The vant Hoff factor of $B a C l_{2}$ at $0.01 M$ concentration is $1.98$ . The percentage of dissociation of $B a C l_{2}$ at this concentration is:

49

69

89

98

100

$B a C l_{2} ⇌ B a^{2 +} + 2 C l^{-}$
Initial 0.01M
At equilibrium $(0.01 - x) M x M 2 x M$
$i = \frac{( 0.01 - x ) + x + 2 x}{0.01}$
$= \frac{0.01 + 2 x}{0.01} = 1.98$
$x = 0.0049$
% $α = \frac{x}{0.01} \times 100 = \frac{0.0049 \times 100}{0.01} = 49$ %

Q. The vant Hoff factor of BaCl2​ at 0.01M concentration is 1.98 . The percentage of dissociation of BaCl2​ at this concentration is:

49

69

89

98

100

BaCl2​⇌Ba2++2Cl− Initial 0.01M At equilibrium (0.01−x)MxM2xM i=0.01(0.01−x)+x+2x​ =0.010.01+2x​=1.98 x=0.0049 % α=0.01x​×100=0.010.0049×100​=49 %

Q. The vant Hoff factor of $B a C l_{2}$ at $0.01 M$ concentration is $1.98$ . The percentage of dissociation of $B a C l_{2}$ at this concentration is:

$B a C l_{2} ⇌ B a^{2 +} + 2 C l^{-}$
Initial 0.01M
At equilibrium $(0.01 - x) M x M 2 x M$
$i = \frac{( 0.01 - x ) + x + 2 x}{0.01}$
$= \frac{0.01 + 2 x}{0.01} = 1.98$
$x = 0.0049$
% $α = \frac{x}{0.01} \times 100 = \frac{0.0049 \times 100}{0.01} = 49$ %