SECTION 3 (Maximum Marks: 16)
This section contains TWO paragraphs
Based on each paragraph, there will be TWO questions
Each question has FOUR options
Text Solution
Verified by Experts200
(i) 2 HCl + NaOH
NaCl + H 2 O
n = 100 x1 = 100 m mole = 0.1 mole
Energy evolved due to neutralization of HCl and NaOH = 0.1 x 57 = 5.7 kJ = 5700 Joule
Energy used to increase temperature of solution = 200 x 4.2 x 5.7 = 4788 Joule
Energy used to increase temperature of calorimeter = 5700 – 4788 = 912 Joule
ms.
t = 912
m.s x 5.7 = 912
ms = 160 Joule/ 0 C [Calorimeter constant]
Energy evolved by neutralization of CH 3 COOH and NaOH
= 200x4.2x5.6x160x5.6=5600 Joule
So energy used in dissociation of 0.1 mole CH3COOH = 5700-5600 = 100 Joule
Enthalpy of dissociation = 1 kJ/mole
(ii) 



pH=4.7
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