[rudiment274]

When 4.21 grams of potassium hydroxide are added to 250 mL of water in a coffee cup calorimeter, the temperature rises by 4.14°C. Assume that the density and specific heat of the dilute aqueous solution are the same as those of H₂O and calculate the molar heat of solution of potassium hydroxide in kJ/mol. C= 4.184 J/g*K
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q_rxn= -q_solution=-m x C x ΔT

So, the mass of the solution is the 4.21 grams of KOH plus the 250 mL of water.
So, that's 4.21 g + (250 mL x 1 g/mL)= 254.21 g

Then we calculate the heat of the reaction by multiplying the mass times the specific heat times the change in temperature.

q_rxn= -254.21 g x 4.184 J/g*°C x 4.14°C = -4247.47 J

Then we convert joules to kJ

Then to find the molar heat of KOH, we take q and divide it by the moles of the solution.

Is my process correct?

[Borek]

Logic looks OK.