A) \[0.46\text{ }M\]
B) \[0.92\text{ }M\]
C) \[0.625\text{ }M\]
D) \[1.25\text{ }M\]
Correct Answer: B
Solution :
Moles of \[Ni{{(N{{O}_{3}})}_{2}}\] in 500 mL of \[2M\,Ni{{(N{{O}_{3}})}_{2}}\] is \[=\frac{2\times 500}{1000}=1mol\] Charge \[=\text{ }96.5\times 18\times 60=104220C\] \[N{{i}^{2+}}+2{{e}^{-}}\xrightarrow{{}}Ni\] \[2\times 96500\]C deposits 1 mole of \[Ni{{(N{{O}_{3}})}_{2}}\] \[\therefore \] 104220 C will deposite \[=\frac{104220}{2\times 96500}=0.54mol\] So, moles of \[Ni{{(N{{O}_{3}})}_{2}}\] left \[=1.0-0.54\text{ }=0.46\]mol Thus, molarity of \[Ni{{(N{{O}_{3}})}_{2}}\] solution \[=2\times 0.46=0.92mol/L\]
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