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 ^0C. (Enthalpy of neutralisation = 57 kJ mol^-1 and Specific he…

Question

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 ^0C. (Enthalpy of neutralisation = 57 kJ mol^-1 and Specific heat of water = 4.2 JK^-1 g^-1) (Neglect heat capacity of flask)

Accepted Answer

Correct answer: 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_2O Bef. 120 40 Aft. 80 0 40m mol ∆_rH = 40m molx (57x10^3) Jm o l =2280J m S∆T = 2280 ⇒1 0 0 0 m L× 1 g mm L×4,2×∆T=2 2 8 0 ∆T= 2 2 8 04.2×1 0^−3 = 2 2 8 0 04 2×1 0^−3 =5 4 2.8 6×1 0^−3 ∆T = 54.286x10^-2K ∆T = 54.286x10^-20C Ans. 54.286 Answer mentioned as 54 (Closest integer)

$When 600 mL of 0.2 M H N O_{3} is mixed with 400 mL of 0.1M NaOH solution in a flask, the rise in temperature of the flask is_______x 1 0^{- 2}^{0} C. (Enthalpy of neutralisation = 57 kJ m o l^{- 1} and Specific heat of water = 4.2 J K^{- 1} g^{- 1}) (Neglect heat capacity of flask)$

Solution

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