Answer:
21.897 °C
Explanation:
Given that:
The heat of solution = 15.2 kJ/mol
Molar mass of ammonium chloride = 53.5 g/mol
So; [tex]\frac{15.2 \frac{kJ}{mol} }{53.5g/mol}[/tex]
= 0.284 kJ/g
Also, the mass of ammonium chloride that was added to the water = 6.314 g
Then the total heat can be calculated as:
0.284 kJ/g × 6.134 g
= 1.742 kJ
= 1742 J
Volume of the water = 65.0 mL = 0.065 L
mass of the water = density × volume
= 1000 g/L × 0.065 L
= 65 g
Total mass can now be = 65 g + 6.314 g
= 71.314 g
The specific heat of water (c) = 4.18 J/g·°C
The heat capacity of the calorimeter = 365 J/°C.
[tex]T_i = 24.5^0C\\T_f = ?\\ \delta T = T_f-24.5^0C[/tex]
Using the formula:
[tex]Q = mc \delta T_{water}+C \delta T_{calorimeter}[/tex]
substituting our values and solving for [tex]T_f[/tex] ; we have:
[tex]1742 = 4.18 *71.314(T_f-24.5^0C)+365*(T_f-24.5^0C)[/tex]
[tex]T_f =[/tex] 21.897 °C
The final temperature of the solution = 21.897 °C
