JEE MAIN - Chemistry (2021 - 27th July Evening Shift - No. 14)
When 400 mL of 0.2 M H2SO4 solution is mixed with 600 mL of 0.1 M NaOH solution, the increase in temperature of the final solution is __________ $$\times$$ 10$$-$$2 K. (Round off to the nearest integer).
[Use : H+ (aq) + OH$$-$$ (aq) $$\to$$ H2O : $$\Delta$$$$\gamma$$H = $$-$$57.1 kJ mol$$-$$1]
Specific heat of H2O = 4.18 J K$$-$$1 g$$-$$1
density of H2O = 1.0 g cm$$-$$3
Assume no change in volume of solution on mixing.
[Use : H+ (aq) + OH$$-$$ (aq) $$\to$$ H2O : $$\Delta$$$$\gamma$$H = $$-$$57.1 kJ mol$$-$$1]
Specific heat of H2O = 4.18 J K$$-$$1 g$$-$$1
density of H2O = 1.0 g cm$$-$$3
Assume no change in volume of solution on mixing.
Answer
82
Explanation
$${n_{{H^ + }}} = {{400 \times 0.2} \over {1000}} \times 2 = 0.16$$
$${n_{O{H^ - }}} = {{600 \times 0.1} \over {1000}} = 0.06$$ (L.R.)
Now, heat liberated from reaction = heat gained by solutions
or, 0.06 $$\times$$ 57.1 $$\times$$ 103
= (1000 $$\times$$ 1.0) $$\times$$ 4.18 $$\times$$ $$\Delta$$T
$$\therefore$$ $$\Delta$$T = 0.8196K
= 81.96 $$\times$$ 10$$-$$2 K $$ \approx $$ 82 $$\times$$ 10$$-$$2 K
$${n_{O{H^ - }}} = {{600 \times 0.1} \over {1000}} = 0.06$$ (L.R.)
Now, heat liberated from reaction = heat gained by solutions
or, 0.06 $$\times$$ 57.1 $$\times$$ 103
= (1000 $$\times$$ 1.0) $$\times$$ 4.18 $$\times$$ $$\Delta$$T
$$\therefore$$ $$\Delta$$T = 0.8196K
= 81.96 $$\times$$ 10$$-$$2 K $$ \approx $$ 82 $$\times$$ 10$$-$$2 K
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