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18 March, 21:37

A 4.00 m solution of KCl was prepared using 1.00 kg of water at 25.0 ∘C. Once the solid had all dissolved, the temperature of the solution was 12.3 ∘C. Calculate the heat of solution, ΔHsoln, of KCl. Assume that the specific heat of the solution is identical to that of water, 4.18 J / (g⋅∘C).

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  1. 18 March, 21:45
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    The heat of solution of KCl is 17.2 kJ

    Explanation:

    Step 1: given data

    We have a 4m solution of KCl prepared in 1kg of water

    ⇒molality = number of moles / kg of solute

    ⇒ in this case it means 4 moles / 1kg of water = 4

    Specific heat of water = 4.18J/g °C

    KCl has a molar mass of 75.55g/mole

    Step 2: calculate mass and change of temperature

    Heat of solution = mass (g) * specific heat (j/g °C) * change of temperature

    mass of KCl we can calculate through te formula mass = moles * Molar mass

    ⇒ m (KCl) = 4 moles * 74.55g/mole = 298.2g

    ⇒ total mass = 298.2g of KCl + 1000g of water = 1298.2g

    Change of temperature = Final temperature T2 - initial temperature T1 = 12.3°C - 25°C = - 12.7 °C ⇔ this means the temperature reduces with 12.7 °C

    Step3 : calculating heat of solution

    heat of solution (q) = mass * specific heat * change in temperature

    q = (1298g) * (4.18J/g °C) * (-12.7) = - 68910 J = - 68.910 kJ

    We will use the positve value of this

    ΔH = q / moles = 68.91 kJ / 4 moles = 17.2 kJ

    The heat of solution of KCl is 17.2 kJ
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