200 mL of 0.2 M HCl is mixed with 300 mL of 0.1 M NaOH. The molar heat of neutralization of this reaction is $-$57.1 kJ. The increase in temperature in $^\circ$C of the system on mixing is x $\times$ 10$-$2. The value of x is ___________. (Nearest integer)
[Given : Specific heat of water = 4.18 J g$-$1 K$-$1, Density of water = 1.00 g cm$-$3]
[Assume no volume change on mixing)
Answer (integer)
82
Solution
$\Rightarrow$ Millimoles of HCl = 200 $\times$ 0.2 = 40<br><br>$\Rightarrow$ Millimoles of NaOH = 300 $\times$ 0.1 = 30<br><br>$\Rightarrow$ Heat released = $\left( {{{30} \over {1000}} \times 57.1 \times 1000} \right)$ = 1713 J<br><br>$\Rightarrow$ Mass of solution = 500 ml $\times$ 1 gm/ml = 500 gm<br><br>$\Rightarrow$ $$\Delta T = {q \over {m \times c}} = {{1713J} \over {500g \times 4.18{J \over {g - K}}}}$$ = 0.8196 K<br><br>= 81.96 $\times$ 10<sup>$-$2</sup> K
About this question
Subject: Chemistry · Chapter: Thermodynamics · Topic: Zeroth and First Law
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