When 400 mL of 0.2 M H₂SO₄ solution is mixed with 600 mL of of 0.1 M NaOH solution the increase in temperature of the final solution is _____ *10⁻² K. (Round off to the Nearest Integer).
[Use : H⁺(aq) + OH⁻(aq) → H₂O : ΔₙH = –57.1kJmol⁻¹
Specific heat of H₂O = 4.18JK⁻¹g⁻¹
Density of H₂O = 1.0gcm⁻³
Assume no change in volume of solution on mixing]
When 400 mL of 0.2 M H₂SO₄ solution is mixed with 600 mL of of 0.1 M NaOH solution the increase in temperature of the final solution is _____ *10⁻² K. (Round off to the Nearest Integer).
[Use : H⁺(aq) + OH⁻(aq) → H₂O : ΔₙH = –57.1kJmol⁻¹
Specific heat of H₂O = 4.18JK⁻¹g⁻¹
Density of H₂O = 1.0gcm⁻³
Assume no change in volume of solution on mixing]
H? SO? + 2NaOH → Na? SO? + 2H? O
Milli mole: 80 60 (limiting reagent)
H? + OH? → H? O
Milli mole: 160 60
Milli mole: 160-60 0 60
Total heat produced = (60/1000) x 57.1 x 1000 = 3426 J
Total heat absorbed = msΔT
ΔT = Totalheat / (m x s)
= 3426J / (1000 x 4.18) = 0.819K
= 81.9 x 10? ² K
Ans. = 82 (the nearest)
Similar Questions for you
CH3COOH + NaOH → CH3COONa + H2O
ΔH = –50.6 kJ/mol
NaOH + SA [HCl] → NaCl + H2O
ΔH = –55.9 kJ/mol
the value of ΔH for ionisation of CH3COOH
⇒ ΔH = +55.9 – 50.6
5.3 kJ/mol
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Chemistry NCERT Exemplar Solutions Class 11th Chapter Eight 2025
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