When 400 mL of 0.2 M H2SO4 solution is mixed with 600 mL of 0.1 M NaOH solution, the increase in temperature of the final solution is __________ $\times$ 10$-$2 K. (Round off to the nearest integer).
[Use : H+ (aq) + OH$-$ (aq) $\to$ H2O : $\Delta$$\gamma$H = $-$57.1 kJ mol$-$1]
Specific heat of H2O = 4.18 J K$-$1 g$-$1
density of H2O = 1.0 g cm$-$3
Assume no change in volume of solution on mixing.
Answer (integer)
82
Solution
${n_{{H^ + }}} = {{400 \times 0.2} \over {1000}} \times 2 = 0.16$<br><br>${n_{O{H^ - }}} = {{600 \times 0.1} \over {1000}} = 0.06$ (L.R.)<br><br>Now, heat liberated from reaction = heat gained by solutions<br><br>or, 0.06 $\times$ 57.1 $\times$ 10<sup>3</sup><br><br>= (1000 $\times$ 1.0) $\times$ 4.18 $\times$ $\Delta$T<br><br>$\therefore$ $\Delta$T = 0.8196K<br><br>= 81.96 $\times$ 10<sup>$-$2</sup> K $\approx$ 82 $\times$ 10<sup>$-$2</sup> K
About this question
Subject: Chemistry · Chapter: Thermodynamics · Topic: Zeroth and First Law
This question is part of PrepWiser's free JEE Main question bank. 110 more solved questions on Thermodynamics are available — start with the harder ones if your accuracy is >70%.