When 400 mL of 0.2 M H_2SO_4 solution is mixed with 600 mL of 0.1 M NaOH solution, the increase in temperature of the final solution is \underline\hspace2cm\times 10 - 2 K. (Round off to the nearest integer). [Use : H^+ (aq) + OH - (aq) \to H_2O : \Delta\gamma H = - 57.1 kJ mol - 1] Specific heat of H_2O = 4.18 J K - 1 g - 1 density of H_2O = 1.0 g cm - 3 Assume no change in volume of solution on mixing.

Thermodynamics PYQ — JEE Main 2021 · 27 Jul · Shift 2 · Q14 with Solution

View written solutionFree

Correct answer: 82

For $H_2SO_4$ : $V = 400\ \text{mL} = 0.4\ \text{L},\quad M = 0.2$ $\text{moles of } H_2SO_4 = 0.4 \times 0.2 = 0.08\ \text{mol}$

Since $H_2SO_4$ gives $2H^+$ per mole, $\text{moles of } H^+ = 2 \times 0.08 = 0.16\ \text{mol}$

For $NaOH$ : $V = 600\ \text{mL} = 0.6\ \text{L},\quad M = 0.1$ $\text{moles of } OH^- = 0.6 \times 0.1 = 0.06\ \text{mol}$

Reaction: $H^+ + OH^- \to H_2O$

We have:

So, $OH^-$ is limiting.

Thus, moles of water formed: $0.06\ \text{mol}$

Given enthalpy of neutralization: $\Delta H = -57.1\ \text{kJ mol}^{-1}$

So heat released is: $q = 0.06 \times 57.1 = 3.426\ \text{kJ}$ $q = 3426\ \text{J}$

Total volume after mixing: $400 + 600 = 1000\ \text{mL}$

Given density $= 1.0\ \text{g cm}^{-3}$ , $\text{mass of solution} = 1000\ \text{g}$

Specific heat: $c = 4.18\ \text{J g}^{-1}\text{K}^{-1}$

Hence, $\Delta T = \frac{q}{mc} = \frac{3426}{1000 \times 4.18}$ $\Delta T = \frac{3426}{4180} \approx 0.8196\ \text{K}$

Temperature rise is of the form: $\underline{\hspace{1cm}} \times 10^{-2}\ \text{K}$

Now, $0.8196\ \text{K} = 81.96 \times 10^{-2}\ \text{K}$

Rounded to nearest integer: $82$

$\boxed{82}$

The derived answer matches the stored correct answer.

$\boxed{82}$

The derived answer matches the stored correct answer.