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JEE Mains · Chemistry · STD 11 - 5. Thermodynamics and thermochemistry

When \(400\, \mathrm{~mL}\) of \(0.2\, \mathrm{M} \,\mathrm{H}_{2} \mathrm{SO}_{4}\) solution is mixed with \(600\, \mathrm{~mL}\) of \(0.1\, \mathrm{M} \,\mathrm{NaOH}\) solution, the increase in temperature of the final solution is \(....\,\times 10^{-2} \,\mathrm{~K}\). (Round off to the Nearest Integer). \(\left[\right.\) Use \(: \mathrm{H}^{+}(\mathrm{aq})+\mathrm{OH}^{-}(\mathrm{aq}) \rightarrow \mathrm{H}_{2} \mathrm{O}: \Delta_{\mathrm{\gamma}} \mathrm{H}=-57.1\, \mathrm{k} \mathrm{J} \,\mathrm{mol}^{-1}\) Specific heat of \(\mathrm{H}_{2} \mathrm{O}=4.18 \mathrm{Jk}^{-} \mathrm{g}^{-}\) density of \(\mathrm{H}_{2} \mathrm{O}=1.0\, \mathrm{~g} \mathrm{~cm}^{-3}\) Assume no change in volume of solution on mixing.]

  1. A \(4\)
  2. B \(2\)
  3. C \(86\)
  4. D \(90\)
Verified Solution

Answer & Solution

Correct Answer

(B) \(2\)

Step-by-step Solution

Detailed explanation

\(\mathrm{n}_{\mathrm{H}^{+}}=\frac{400 \times 0.2}{1000} \times 2=0.16\) \(\mathrm{n}_{\mathrm{OH}^{-}}=\frac{600 \times 0.1}{1000}=0.06\, \text { (L.R.) }\) Now, heat liberated from reaction \(=\) heat gained by solutions or, \(0.06 \times 57.1 \times 10^{3}\)…
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