When 400ml of 0.2M H2SO4 solution is


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).

$\left[\mathrm{Use}: \mathrm{H}^{+}(\mathrm{aq})+\mathrm{OH}^{-}(\mathrm{aq}) \rightarrow \mathrm{H}_{2} \mathrm{O}:\right.$ $\left.\Delta_{\gamma} \mathrm{H}=-57.1 \mathrm{~kJ} \mathrm{~mol}^{-1}\right]$

Specific heat of $\mathrm{H}_{2} \mathrm{O}=4.18 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~g}^{-1}$

density of $\mathrm{H}_{2} \mathrm{O}=1.0 \mathrm{~g} \mathrm{~cm}^{-3}$

Assume no change in volume of solution on mixing.


$\mathrm{n}_{\mathrm{H}^{+}}=\frac{400 \times 0.2}{1000} \times 2=0.16$

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

Now, heat liberated from reaction

$=$ heat gained by solutions

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

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

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

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

Leave a comment