Answer:
Equation correctly showing the heat of solution
![KNO_3(s)+35.2 kJ \rightarrow K^+(aq) + NO_(3)^-(aq)](https://img.qammunity.org/2020/formulas/chemistry/college/lknyyu7zi3n6diqvna4v855vq63nkajte9.png)
Step-by-step explanation:
Mass of aqueous solution = m = 100 g
Specific heat of solution = c = 4.18 J/gºC
Change in temperature =
![\Delta T=T_f-T_i](https://img.qammunity.org/2020/formulas/physics/high-school/b6hlrtgleochfjy577x28vmaiagtmip0mu.png)
ΔT = 21.6ºC - 30.0ºC = -8.4ºC
Heat lost by the solution = Q
![Q = mc\Delta T](https://img.qammunity.org/2020/formulas/physics/college/7yajqh5fmxg5g85kia5s3lr6sbs5vi6l4m.png)
![Q=100 g* 4.18 J/g^oC* (-8.4^oC)](https://img.qammunity.org/2020/formulas/chemistry/college/r0hcoulw1xk648g68mjv3qx01ubbvy7tuw.png)
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=
![(10.1 g)/(101 g/mol)=0.1 mol](https://img.qammunity.org/2020/formulas/chemistry/college/tpo5gkjifjbii5homjtza4zf6rmpuz7nbh.png)
![KNO_3(s)\rightarrow K^+(aq) + NO_(3)^-(aq)](https://img.qammunity.org/2020/formulas/chemistry/college/vmjhr4ol8arfod3iknr9fxtqssu1gbhvzx.png)
The heat of solution =
![(Q')/(n)=(3.5112 kJ)/(0.1 mol)=35.1 kJ/mol](https://img.qammunity.org/2020/formulas/chemistry/college/u5y8ocoe7bxscjvknlszckv28crv9ntqjb.png)
So, the equation correctly showing the heat of solution
![KNO_3(s)+35.2 kJ \rightarrow K^+(aq) + NO_(3)^-(aq)](https://img.qammunity.org/2020/formulas/chemistry/college/lknyyu7zi3n6diqvna4v855vq63nkajte9.png)