When 400mL of 0 . 2M H _ 2 SO _ 4 solution is mixed with 600mL of 0...

When $400mL of 0 . 2M{H}_{2}{SO}_{4}$ solution is mixed with $600mL of 0 . 1MNaOH$ solution, the increase in temperature of the final solution is $— \times {10}^{- 2}K$ . (Round off to the nearest integer).

[Use : ${H}^{+} ( aq ) + {OH}^{-} ( aq ) \to {H}_{2}O:{\Delta}_{\gamma}H = - 57 . 1kJ{mol}^{- 1}] Specific heat of$ {H}_{2}O = 4 . 18J{K}^{- 1}{g}^{- 1} $, density of$ {H}_{2}O = 1 . 0g{cm}^{- 3} $Assume no change in volume of solution on mixing.$

Correct answer

Step-by-step solution

$n{H}^{+} = \dfrac{400 \times 0 . 2}{1000} \times 2 = 0 . 16$

$n{OH}^{-} = \dfrac{600 \times 0 . 1}{1000} = 0 . 06 ( LimitingReagent )$

Now we can apply, $q = ms ∆ T$

$0 . 06 \times 57 . 1 \times {10}^{3}$

$= ( 1000 \times 1 . 0 ) \times 4 . 18 \times \Delta T$

$∴ \Delta T = 0 . 8196K$

$= 81 . 96 \times {10}^{- 2}K \approx 82 \times {10}^{- 2}K$

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