JEE Main 2022 — Chemistry Physical Chemistry

2022integerhard

When $600\mathrm{mL}$ of $0.2M{\mathrm{HNO}}_{3}$ is mixed with $400\mathrm{mL}$ of $0.1M\mathrm{NaOH}$ solution in a flask, the rise in temperature of the flask is_____ $\times {10}^{-2}^{\circ}C$

(Enthalpy of neutralisation $=57\mathrm{kJ}{\mathrm{mol}}^{-1}$ and Specific heat of water $=4.2{\mathrm{JK}}^{-1}{g}^{-1}$ ) (Neglect heat capacity of flask)

Official previous-year question

Held on 29 Jul 2022 · Verified 6 Jul 2026.

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Solution

Number of moles of given acid and base are

$\begin{matrix}{\mathrm{HNO}}_{3}=\frac{600}{1000}\times 0.2M=0.12\mathrm{moles} & \end{matrix} \mathrm{NaOH}=\frac{400}{1000}\times 0.1M=0.04\mathrm{moles}$

Now neutralisation reaction between them will be

${\mathrm{HNO}}_{3}+\mathrm{NaOH}\rightarrow {\mathrm{NaNO}}_{3}+{H}_{2}O 0.12\mathrm{mole}0.04\mathrm{mole} 0.08\mathrm{mole}0\mathrm{mole}0.04\mathrm{mole}$

Heat relesed during this reaction will be

${\Delta }_{r}H=0.04\times (57\times {10}^{3})J$

$=2280J$

Now temperature increased can be calculated by

$mS\Delta T=2280J$

$\Rightarrow 1000\mathrm{mL}\times \frac{1\mathrm{gm}}{\mathrm{mL}}\times 4.2\times \Delta T=2280$

$\Delta T=\frac{2280}{4.2}\times {10}^{-3}$

$=542.86\times {10}^{-3}$

$\Delta T=54.286\times {10}^{-2}^{\circ}C$

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JEE Main 2022 — Chemistry Physical Chemistry

2022integerhard

When $600\mathrm{mL}$ of $0.2M{\mathrm{HNO}}_{3}$ is mixed with $400\mathrm{mL}$ of $0.1M\mathrm{NaOH}$ solution in a flask, the rise in temperature of the flask is_____ $\times {10}^{-2}^{\circ}C$

(Enthalpy of neutralisation $=57\mathrm{kJ}{\mathrm{mol}}^{-1}$ and Specific heat of water $=4.2{\mathrm{JK}}^{-1}{g}^{-1}$ ) (Neglect heat capacity of flask)

Official previous-year question

Held on 29 Jul 2022 · Verified 6 Jul 2026.

Did you get this right?

Solution

Number of moles of given acid and base are

$\begin{matrix}{\mathrm{HNO}}_{3}=\frac{600}{1000}\times 0.2M=0.12\mathrm{moles} & \end{matrix} \mathrm{NaOH}=\frac{400}{1000}\times 0.1M=0.04\mathrm{moles}$

Now neutralisation reaction between them will be

${\mathrm{HNO}}_{3}+\mathrm{NaOH}\rightarrow {\mathrm{NaNO}}_{3}+{H}_{2}O 0.12\mathrm{mole}0.04\mathrm{mole} 0.08\mathrm{mole}0\mathrm{mole}0.04\mathrm{mole}$

Heat relesed during this reaction will be

${\Delta }_{r}H=0.04\times (57\times {10}^{3})J$

$=2280J$

Now temperature increased can be calculated by

$mS\Delta T=2280J$

$\Rightarrow 1000\mathrm{mL}\times \frac{1\mathrm{gm}}{\mathrm{mL}}\times 4.2\times \Delta T=2280$

$\Delta T=\frac{2280}{4.2}\times {10}^{-3}$

$=542.86\times {10}^{-3}$

$\Delta T=54.286\times {10}^{-2}^{\circ}C$

Your note

JEE Main Physical Chemistry in other years

Keep practicing Physical Chemistry

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JEE Main 2022 — Chemistry Physical Chemistry

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Solution

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JEE Main Physical Chemistry in other years

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JEE Main 2022 — Chemistry Physical Chemistry

Did you get this right?

Solution

Your note

JEE Main Physical Chemistry in other years

More JEE Main Physical Chemistry questions

Other Chemistry topics for JEE Main

Organic Chemistry

Inorganic Chemistry

Keep practicing Physical Chemistry