When 600 mL of 0.2 M HNO 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 °C. (Enthalpy of neutralisation = 57 kJ mol –1 and Specific heat of water = 4.2 JK –1 g –1 ) (Neglect heat capacity of flask)

Question

When 600 mL of 0.2 M HNO 3  is mixed with 400 mL of 0.1 M NaOH solution in a flask, the rise in temperature of the flask is ________ &times; 10 &ndash;2  &deg;C. (Enthalpy of neutralisation = 57 kJ mol &ndash;1  and Specific heat of water = 4.2 JK &ndash;1  g &ndash;1 ) (Neglect heat capacity of flask)

cdquestions Admin · Accepted Answer

$\text{H}^+ + \text{OH}^- \rightarrow \text{H}_2\text{O}$ m moles 120 40 – 80 – 40 Heat liberated from reaction = 40 × 10 –3 × 57 × 10 3  J …(1) Heat gained by solution =  $mC\Delta T$ Mass of solution  $m = V × d = 1000 × 1 = 1000 g$ Heat gained by solution  $= 1000 × 4.2 × ΔT …(2)$ From eq (1) and eq (2) Heat liberated = Heat gained $40 × 10–3 × 57 × 103 = 1000 × 4.2 × ΔT$ $ΔT = 54 ×$  10 –2   $°C$ $ΔT = 54°C$  (Rounded off to the nearest integer) So, the answer is 54°C.

When 600 mL of 0.2 M HNO₃ is mixed with 400 mL of 0.1 M NaOH solution in a flask, the rise in temperature of the flask is ________ × 10^–2°C.
(Enthalpy of neutralisation = 57 kJ mol^–1 and Specific heat of water = 4.2 JK^–1 g^–1)
(Neglect heat capacity of flask)

Correct Answer: 54

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When 600 mL of 0.2 M HNO3 is mixed with 400 mL of 0.1 M NaOH solution in a flask, the rise in temperature of the flask is ________ × 10–2 °C.(Enthalpy of neutralisation = 57 kJ mol–1 and Specific heat of water = 4.2 JK–1 g–1)(Neglect heat capacity of flask)