Suppose 5 ml of 1.5 M ethylene diamine is added to 50 mlof 0.15 M Ni 2+ solution

Question

Explanation / Answer

Q = mct

Assume density = 1g/dm3

1 ml = 1 dm3

m = 50+5 = 55g

Q = (55)(c)(1.1) = 60.5 x c

H = Q/No.of moles

No of moles = C x V

No of moles of ethylene diamine = 1.5 x 5 = 7.5 moles

No of moles of Ni2+ = 50 x 0.15 = 7.5 moles

Total number of moles = 7.5 + 7.5 = 15 moles

Therefore,

H = 60.5 x c / 15 = 4.0333c J/mol

= 4.0333 x c x 10-3 KJ/mol

Substitute c value, then final answer can be obtained.

(since c value is not given in the question )

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