Question

When 400 mL of 0.2M H₂SO₄ solution is mixed with 600 mL of 0.1 M NaOH solution, the increase in temperature of the final solution is ..... x 10^–2 K. (Round off to the nearest integer).

[Use : H⁺ (aq) + OH^– (aq) → H₂O : 

△_γH = – 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 1

Answer is 82

Now, heat liberated from reaction 

= heat gained by solutions 

or, 0.06 x 57.1 x 10³ 

= (1000 x 1.0) x 4 .18 x △T 

△T = 0.8196 K 

= 81.96 x 10^–2 K = 82 x 10^–2 K

Comments

Why is mass taken equal to 1000 ?

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M = density x volume
= 1.0 x 1000 m2
M = 1000 g