`100 cm^(3)` of `0.05N 0.5 HCI` solution at `299.95 K` was mixed with `100 cm^(3) 0.5 N NaOH` solution at `299.75 K` in a thermos flask. The final temperature was found to be `302.65K`. Calculate the enthalpy of neutralisation of `HCI`. Water equivalent of thermos flask is `44 g`.

`100 cm^(3)` of `0.05N 0.5 HCI` solution at `299.95 K` was mixed with

1.	`100 cm^(3)` of `0.05N 0.5 HCI` solution at `299.95 K` was mixed with `100 cm^(3) 0.5 N NaOH` solution at `299.75 K` in a thermos flask. The final temperature was found to be `302.65K`. Calculate the enthalpy of neutralisation of `HCI`. Water equivalent of thermos flask is `44 g`.
Answer» The initial average temperature of the acid and the base `=(299.95+299.75)/(2) = 299.85K` Rise in temperature `= (302.65-299.85) = 2.80K` Heat evolved during neutralisation `= (100+100+44) xx 4.184 xx 2.8 = 2858.5J` `:.` Ethalpy of neutralisation `=- (2858.5)/(100) xx 1000xx(1)/(0.50)` `=- 57.17 kJ`

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`100 cm^(3)` of `0.05N 0.5 HCI` solution at `299.95 K` was mixed with `100 cm^(3) 0.5 N NaOH` solution at `299.75 K` in a thermos flask. The final temperature was found to be `302.65K`. Calculate the enthalpy of neutralisation of `HCI`. Water equivalent of thermos flask is `44 g`.

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