1.

Experiment shows that Nickel oxide has the formula Ni096 O1.00 . What fraction of Nickel exists as of Ni2+ and Ni3+ ions ?

Answer»

Let the number of Ni2+ . Then the number of Ni3+ ion will be = (0.96 – x). 

Total number of cation, = 2 x + 3 (0.96 – x) 

= 2x + 2.88 – 3x 

= ( – x) + 2.88

Number of anions O2- ( – 2) x 1 = -2. Number of cations = Number of anions – x + 2.88 2

– x = – 2.88 + 2

x = 0.88

 % of Ni as Ni2 = \(\frac{0.08}{0.96}\times 100\) = 91.66% 

Number of Ni3+ ion will be = 0.96 – x 

= 0.96 – 0.88 = 0.08 

% of Ni as Ni3+ =   \(\frac{0.08}{0.96}\times 100\)  = 8.3 %



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