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  3. Chemistry
  4. 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 ° C of the system on mixing is x × 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)

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Q. $200\, mL$ 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)

JEE MainJEE Main 2021Thermodynamics

Solution:

$\Rightarrow $ Millimoles of $HCl =200 \times 0.2=40$
$\Rightarrow $ Millimoles of $NaOH =300 \times 0.1=30$
$\Rightarrow $ Heat released $=\left(\frac{30}{1000} \times 57.1 \times 1000\right)=1713\, J$
$\Rightarrow $ Mass of solution $=500 \, ml \times 1\, gm / ml =500 \, gm$
$\Rightarrow \Delta T =\frac{ q }{ m \times C }=\frac{1713 J }{500 g \times 4.18 \frac{ J }{ g - K }}=0.8196 \, K$
$=81.96 \times 10^{-2} \, K $