when a 4.00 gram sample of KBr is dissolved in water in a calorimeter that has a total heat capacity of 3.185kJ*K

^{-1}, the temperature decreases by 0.210 K. Calculate the molar heat of solution of KBr.
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Rock Hill, SC

Harley,

This is a question about calorimetry. We are given a temperature change and the total heat capacity of the calorimeter. We are also given the amount ( in grams) of KBr being dissolved. Our goal is to calculate the enthalpy change
per mole of KBr.

First, we will calculate the enthalpy change and then convert that heat into a molar quantity.

To solve, we will use the equation

So, using the values given

ΔH = -(3.185 kJ K^{-1}) x (-0.210 K) Notice that ΔT is negative because we are told the temperature decreased.

ΔH = 0.669 kJ, but we need to convert to a molar quantity, so

(0.669 kJ/4.00 g KBr) x (119.0 g KBr/1 mol KBr)

Grams cancel and we get 0.669 kJ/ 0.0336 mol KBr = 19.9 kJ/mol

I hope this helps!

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