Published by:
CGP EDU Academic Team
Published on: August 13, 2026
When 600 mL of 0.2 M HNO 3 is mixed with 400 mL of 0.1M NaOH solution in a flask, the rise in temperature of the flask is_______x 10 -2 0 C. (Enthalpy of neutralisation = 57 kJ mol -1 and Specific heat of water = 4.2 JK -1 g -1 )
(Neglect heat capacity of flask)
Text Solution
Verified by ExpertsThe correct answer is:
54
(54)
HNO 3 NaOH
600 mL x 0.2 M 400 mL x 0.1 M
= 120 m mol = 40 m mol
HNO 3 + NaOH 
NaNO 3 + H 2 O
Bef. 120 40
Aft. 80 0 40m mol
r H = 40m molx (57x10 3 ) 
=2280J
m S 
T = 2280
T = 54.286x10 -2 K
T = 54.286x10 -20 C
Ans. 54.286
Answer mentioned as 54 (Closest integer)
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