ThermochemistryHard

Question

A solution was prepared by dissolving 7.45 g of KCl in 200 g of H2O in a calorimeter with a water equivalent of 25 g. The temperature of the water had reduced from 28oC to 25oC. The value of ΔH (in kJ/ mol) for dissolving KCl in water is (Specific heat capacity of water is 4.2 J/K-g)

Options

A.+2.52
B.+2.835
C.+25.2
D.+28.35

Solution

Heat absorbed in solubility = Heat released from solution

$= m.s.\Delta T $$${= (200 + 25) \times 4.2 \times 3 = 2835\text{ J} }{\therefore\Delta H = \frac{2835}{7.45} \times 74.5 = 28350\text{ J}}$$

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