Question

a chemist dissolved an 11.9-g sample of koh in 100.0 grams of water in a coffee cup calorimeter. when she did so, the water temperature increased by 26.0 ∘c. based on this, how much heat energy was required to dissolve the sample of koh? assume the specific heat of the solution is 4.184 j/g⋅°c.

Answer 1

The heat of solution is -51.8 kJ/mol

What is the heat of solution?

We know that in a calorimeter, there is no loss or gain of energy. It is a good example of a closed system.

Number of moles of KOH =  11.9-g/56 g/mol = 0.21 moles

Temperature rise = 26.0 ∘c

Mass of the water = 100.0 grams

Heat capacity =  4.184 j/g⋅°c

Then;

ΔH = mcθ

ΔH = 100g * 4.184 j/g⋅°c * 26.0 ∘c = 10.88 kJ

Heat of solution = -(10.88 kJ/ 0.21 moles) = -51.8 kJ/mol

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