Question:
The vant Hoff factor of $ BaC{{l}_{2}} $ at $ 0.01 \,M $ concentration is $ 1.98 $ . The percentage of dissociation of $ BaC{{l}_{2}} $ at this concentration is:
The vant Hoff factor of $ BaC{{l}_{2}} $ at $ 0.01 \,M $ concentration is $ 1.98 $ . The percentage of dissociation of $ BaC{{l}_{2}} $ at this concentration is:
Updated On: Jun 6, 2022
- 49
- 69
- 89
- 98
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The Correct Option is A
Solution and Explanation
$ BaC{{l}_{2}} \rightleftharpoons B{{a}^{2+}}+2C{{l}^{-}} $
Initial 0.01M
At equilibrium $ (0.01-x)MxM2xM $
$ i=\frac{(0.01-x)+x+2x}{0.01} $
$ =\frac{0.01+2x}{0.01}=1.98 $
$ x=0.0049 $
% $ \,\alpha=\frac{x}{0.01}\times 100=\frac{0.0049\times 100}{0.01}=49 $ %
Initial 0.01M
At equilibrium $ (0.01-x)MxM2xM $
$ i=\frac{(0.01-x)+x+2x}{0.01} $
$ =\frac{0.01+2x}{0.01}=1.98 $
$ x=0.0049 $
% $ \,\alpha=\frac{x}{0.01}\times 100=\frac{0.0049\times 100}{0.01}=49 $ %
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Rate of the reaction is given by,
Rate = −d[A]/ dt=+d[B]/ dt
Where, [A] → concentration of reactant A
[B] → concentration of product B
(-) A negative sign indicates a decrease in the concentration of A with time.
(+) A positive sign indicates an increase in the concentration of B with time.
Factors Determining the Rate of a Reaction:
There are certain factors that determine the rate of a reaction:
- Temperature
- Catalyst
- Reactant Concentration
- Chemical nature of Reactant
- Reactant Subdivision rate