A) 25%
B) 50%
C) 75%
D) 80%
Correct Answer: C
Solution :
\[\begin{align} & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,Ca{{(N{{O}_{3}})}_{2}}\rightleftharpoons C{{a}^{2+}}+2NO_{3}^{-} \\ & \text{Initially}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\text{1}\,\text{mol}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\text{0}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\text{0} \\ & \text{At }\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,(1-x)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,x\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,2x \\ & \text{equilibrium} \\ \end{align}\] Total number of particles (moles) at equilibrium \[\,\,\,=(1-x)\,+x+2x\,\,\,\] \[=(1+2x)\] \[\text{Now,}\frac{\text{calculated}\,\text{molecular}\,\text{mass}}{\text{normal}\,\text{molecular}\,\text{mass}}\] \[\text{=}\frac{\text{number}\,\text{of}\,\text{particles}\,\text{after}\,\text{dissociation}}{\text{number}\,\text{of}\,\text{particles}\,\text{before}\,\text{dissociation}}\] \[\frac{164}{65.4}=\frac{1+2x}{1}\] or \[\frac{164-65.4}{65.4}=2x\] or \[x=\frac{98.6}{65.4\times 2}=0.7538\] or degree of dissociation of \[\text{Ca(N}{{\text{O}}_{\text{3}}}{{\text{)}}_{\text{2}}}\] \[=0.7538\times 100=75.38%\]
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