Question

A coffee cup calorimeter initially contains 125g of water at
24.2^{o}C. 10.5g of potassium bromide also at
24.2^{o}C is added to the water. After the KBr dissolves
the final temperature is 21.1^{o}C. Calculate the enthalpy
change for dissolving the salt in J/g and kJ/mol. Assume specific
heat of solution is 4.18J/g^{o}C.

Answer #1

Sol :-

Total mass of KBr solution (m) = 125 g + 10.5 g = 135.5 g

Heat evolve (q) = Mass of solution (m) x Heat capacity of water (C) x Change in temperature (ΔT)

= (135.5 g).(4.18
J/g.^{0}C).(21.1-24.2)^{0}C

= - 1755.809 J

Now, enthalpy change (ΔH) per gram of KBr solution = -1755.809 J/135.5 g

= **-12.958 J/g**

As, Number of moles of KBr = Given mass of KBr/Gram molar mass of KBr

= 10.5 g / 119.002 g/mol

= 0.0882 mol

So,

Enthalpy change (ΔH) per mole of KBr solution = - 1755.809 J / 0.0882 mol

= - 19907.13 J/mol

= **- 19.907 KJ/mol**

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