Answer:
Equation correctly showing the heat of solution
[tex]KNO_3(s)+35.2 kJ \rightarrow K^+(aq) + NO_{3}^-(aq)[/tex]
Explanation:
Mass of aqueous solution = m = 100 g
Specific heat of solution = c = 4.18 J/gºC
Change in temperature = [tex]\Delta T=T_f-T_i[/tex]
ΔT = 21.6ºC - 30.0ºC = -8.4ºC
Heat lost by the solution = Q
[tex]Q = mc\Delta T[/tex]
[tex]Q=100 g\times 4.18 J/g^oC\times (-8.4^oC)[/tex]
Q = -3,511.2 J ≈ -3.51 kJ
Heat absorbed by potassium nitrate when solution in formed; Q'
Q' = -Q = 3.51 kJ
Moles of potassium nitrate , n= [tex]\frac{10.1 g}{101 g/mol}=0.1 mol[/tex]
[tex]KNO_3(s)\rightarrow K^+(aq) + NO_{3}^-(aq)[/tex]
The heat of solution =
[tex]\frac{Q'}{n}=\frac{3.5112 kJ}{0.1 mol}=35.1 kJ/mol[/tex]
So, the equation correctly showing the heat of solution
[tex]KNO_3(s)+35.2 kJ \rightarrow K^+(aq) + NO_{3}^-(aq)[/tex]