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]

0 5 Views | Posted a month ago
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    Answered by

    alok kumar singh | Contributor-Level 10

    a month ago

    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)

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