When 600 mL of $0.2 \mathrm{M} \mathrm{HNO}_{3}$ is mixed with 400 mL of 0.1 M NaOH solution in a flask, the rise in temperature of the flask is ______ $\times 10^{-2}$ $^{0}\mathrm{C}$. (Enthalpy of neutralisation $=57 \mathrm{~kJ} \mathrm{mo1}^{-1}$ and Specific heat of water $=4.2 \mathrm{JK}^{-1} \mathrm{~g}^{-1}$) (Neglect heat capacity of flask)

JEE Main · 2022 · 29 Jul 2022 · Shift 1 · Chemical Thermodynamics

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 1 0^{- 2}

^{0} C

.
(Enthalpy of neutralisation

= 57 kJ mo1^{- 1}

and Specific heat of water

= 4.2 JK^{- 1} g^{- 1}

)
(Neglect heat capacity of flask)

Answer: 54

Moles of

HNO X_{3}

0.6 \times 0.2 = 0.12

mol = 120 mmol

Moles of NaOH =

0.4 \times 0.1 = 0.04

mol = 40 mmol

NaOH is limiting reagent. Heat released:

q = 40 \times 1 0^{- 3} \times 57 \times 1 0^{3} = 2280 J

Total volume = 1000 mL = 1000 g (assuming density = 1 g/mL)

Using

q = m s Δ T

2280 = 1000 \times 4.2 \times Δ T

Δ T = \frac{2280}{4200} = 0.543 K = 54.3 \times 1 0^{- 2} °C

Nearest integer: 54

JEE Main · 2022 · 29 Jul 2022 · Shift 1 · Integer answer