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
Question:

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)

Solution:

$\Rightarrow$ 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 $=500 \mathrm{ml} \times 1 \mathrm{gm} / \mathrm{ml}=500 \mathrm{gm}$

$\Rightarrow \Delta \mathrm{T}=\frac{\mathrm{q}}{\mathrm{m} \times \mathrm{C}}=\frac{1713 \mathrm{~J}}{500 \mathrm{~g} \times 4.18 \frac{\mathrm{J}}{\mathrm{g}-\mathrm{K}}}=0.8196 \mathrm{~K}$

$=81.96 \times 10^{-2} \mathrm{~K}$

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