Respuesta :
Answer:
32.04°C will be the final temperature of the solution.
Explanation:
Moles of potassium chloride = 0.200 mol
MAs sof KCl= 0.200 mol × 74.5 g/mol= 14.9 g
Enthalpy of solvation of potassium nitrate =
[tex]\Delta H_{solv}=17.1 kJ/mol[/tex]
Energy released when 0.200 moles of KCl is dissolved in water = Q
[tex]Q=17.1kJ/mol\times 0.200 mol=3.42 kJ=3420 J[/tex]
(1 kJ = 1000 J)
Heat released on dissolving 0.200 moles of KCl is equal to heat absorbed by water = Q
Mass of solution , m= 80.0 g +14.9 g = 94.9 g
Specific heat of water = c = 4.184 J/g°C
Initial temperature of the water = [tex]T_1=23.4^oC[/tex]
Final temperature of the water = [tex]T_2=?[/tex]
[tex]Q=m\times c\times (T_2-T_1)[/tex]
[tex]3420 J=94.9g\times 4.184 J/g^oC\times (T_2-23.4^oC)[/tex]
[tex]T_2=32.04^oC[/tex]
32.04°C will be the final temperature of the solution.
