MCQ
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.]

  • A
    $4$
  • $2$
  • C
    $86$
  • D
    $90$

Answer

Correct option: B.
$2$
b
$\mathrm{n}_{\mathrm{H}^{+}}=\frac{400 \times 0.2}{1000} \times 2=0.16$

$\mathrm{n}_{\mathrm{OH}^{-}}=\frac{600 \times 0.1}{1000}=0.06\, \text { (L.R.) }$

Now, heat liberated from reaction

$=$ heat gained by solutions

or, $0.06 \times 57.1 \times 10^{3}$

$=(1000 \times 1.0) \times 4.18 \times \Delta \mathrm{T}$

$\therefore \Delta \mathrm{T}=0.8196\, \mathrm{~K}$

$=81.96 \times 10^{-2}\, \mathrm{~K} \approx 82 \times 10^{-2} \,\mathrm{~K}$

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