1. Tardigrade
  2. Question
  3. Chemistry
  4. 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 0 C. (Enthalpy of neutralisation =57 kJ mol -1 and Specific heat of water =4.2 JK -1 g -1 ) (Neglect heat capacity of flask)

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Q. 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 ___$\times 10^{-2}{ }^0 C$.
(Enthalpy of neutralisation $=57 \, kJ \, mol ^{-1}$ and Specific heat of water $=4.2 \, JK ^{-1}\, g ^{-1}$ )
(Neglect heat capacity of flask)

JEE MainJEE Main 2022Thermodynamics

Solution:

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$ \Delta_{ r } H =40 \,m \,mol \times\left(57 \times 10^3\right) \frac{ J }{ mol } $
$=40 \times 10^{-3}\, mol \times 57 \times 10^3 \frac{ J }{ mol }$
$= 2280 J $
$ m S \Delta T =2280$
$ \Rightarrow 1000 mL \times \frac{1 gm }{ mL } \times 4,2 \times \Delta T =2280
$
$\Delta T =\frac{2280}{4.2} \times 10^{-3} $
$=\frac{22800}{42} \times 10^{-3}$
$= 542.86 \times 10^{-3}$
$\Delta T =54.286 \times 10^{-2} K$
$\Delta T =54.286 \times 10^{-20} C$
Ans. $54.286$
Answer mentioned as 54 (Closest integer)