JEE MAIN - Chemistry (2022 - 29th July Morning Shift - No. 18)
When 600 mL of 0.2 M HNO3 is mixed with $$400 \mathrm{~mL}$$ of 0.1 M NaOH solution in a flask, the rise in temperature of the flask is ___________ $$\times 10^{-2}{ }\,^{\circ} \mathrm{C}$$.
(Enthalpy of neutralisation $$=57 \mathrm{~kJ} \mathrm{~mol}^{-1}$$ and Specific heat of water $$=4.2 \,\mathrm{JK}^{-1} \mathrm{~g}^{-1}$$) (Neglect heat capacity of flask)
Answer
54
Explanation
HNO3
600 mL × 0.2 M = 120 m mol
NaOH
400 mL × 0.1 M = 40 m mol
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Heat liberated from reaction
$$ =40 \times 10^{-3} \times 57 \times 10^{3} \mathrm{~J} $$
Heat gained by solution $=m C \Delta T$
$$ \begin{aligned} \mathrm{m}=\text { mass of solution }=\mathrm{V} \times \mathrm{d} &=1000 \times 1 \\\\ &=1000 \mathrm{~g} \end{aligned} $$
Heat gained by solution $=1000 \times 4.2 \times \Delta \mathrm{T} \ldots(2)$
From (1) and (2)
Heat liberated $=$ Heat gained
$40 \times 10^{-3} \times 57 \times 10^{3}=1000 \times 4.2 \times \Delta T$
$\Delta T=54 \times 10^{-2}{ }^{\circ} \mathrm{C}$
(Rounded off to the nearest integer)
600 mL × 0.2 M = 120 m mol
NaOH
400 mL × 0.1 M = 40 m mol
Heat liberated from reaction
$$ =40 \times 10^{-3} \times 57 \times 10^{3} \mathrm{~J} $$
Heat gained by solution $=m C \Delta T$
$$ \begin{aligned} \mathrm{m}=\text { mass of solution }=\mathrm{V} \times \mathrm{d} &=1000 \times 1 \\\\ &=1000 \mathrm{~g} \end{aligned} $$
Heat gained by solution $=1000 \times 4.2 \times \Delta \mathrm{T} \ldots(2)$
From (1) and (2)
Heat liberated $=$ Heat gained
$40 \times 10^{-3} \times 57 \times 10^{3}=1000 \times 4.2 \times \Delta T$
$\Delta T=54 \times 10^{-2}{ }^{\circ} \mathrm{C}$
(Rounded off to the nearest integer)
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