JEE Mains · Chemistry · STD 11 - 5. Thermodynamics and thermochemistry
\(200\, \mathrm{~mL}\) of \(0.2\, \mathrm{M} \mathrm{HCl}\) is mixed with \(300\, \mathrm{~mL}\) of \(0.1\, \mathrm{M} \mathrm{NaOH}\). The molar heat of neutralization of this reaction is \(-57.1 \,\mathrm{~kJ}\). The increase in temperature in \({ }^{\circ} \mathrm{C}\) of the system on mixing is \(\mathrm{x} \times 10^{-2}\). The value of \(\mathrm{x}\) 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)
- A \(12\)
- B \(125\)
- C \(82\)
- D \(74\)
Answer & Solution
Correct Answer
(C) \(82\)
Step-by-step Solution
Detailed explanation
\(\Rightarrow \text { Millimoles of } \mathrm{HCl}=200 \times 0.2=40\) \(\Rightarrow \text { Millimoles of } \mathrm{NaOH}=300 \times 0.1=30\) \(\Rightarrow \text { Heat released }=\left(\frac{30}{1000} \times 57.1 \times 1000\right)=1713 \,\mathrm{~J}\)…
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