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
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 ${ }^{\circ} \mathrm{C}$ of the system on mixing is $x \times 10^{-2}$.
The value of x is ________ . (Nearest integer)
[Given : Specific heat of water $=4.18 \mathrm{~J} \mathrm{~g}^{-1} \mathrm{~K}^{-1}$
Density of water $=1.00 \mathrm{~g} \mathrm{~cm}^{-3}$ ]
(Assume no volume change on mixing)
$\Rightarrow$ Millimoles of $\mathrm{HCl}=200 \times 0.2=40$
$\Rightarrow$ Millimoles of $\mathrm{NaOH}=300 \times 0.1=30$
$\Rightarrow$ Heat released $=\left(\frac{30}{1000} \times 57.1 \times 1000\right)=1713 \mathrm{~J}$
$\Rightarrow$ Mass of solution $=500 \mathrm{ml} \times 1 \mathrm{gm} / \mathrm{ml}=500 \mathrm{gm}$
$\Rightarrow \Delta \mathrm{T}=\frac{\mathrm{q}}{\mathrm{m} \times \mathrm{C}}=\frac{1713 \mathrm{~J}}{500 \mathrm{~g} \times 4.18 \frac{\mathrm{J}}{\mathrm{g}-\mathrm{K}}}=0.8196 \mathrm{~K}$
$=81.96 \times 10^{-2} \mathrm{~K}$