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JEE Main 2022 · 29 Jul · Shift 1, Chemistry question 84

When 600 mL of 0.2 M HNO3_3 is mixed with 400 mL of 0.1 M NaOH solution in a flask, the rise in temperature of the flask is _______ ×10−2\times10^{-2} °C.

(Enthalpy of neutralisation = 57 kJ mol−1^{-1} and Specific heat of water = 4.2 J K−1^{-1} g−1^{-1})

(Neglect heat capacity of flask)

Answer and worked solution

Official answer: 54

Neutralised. NaOH 0.04 mol is limiting: heat 0.04×57000=22800.04\times57000=2280 J.

Rise in 1000 g of solution: 22801000×4.2=0.543\frac{2280}{1000\times4.2}=0.543 °C, i.e. 54×10−254\times10^{-2}.

Answer: 54

Topics:Calorimetry
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