The heat of solution of ammonium chloride is 15.2 kJ/mol. If a 6.134 g sample of NH₄Cl is added to 65.0 mL of water in a calorimeter at 24.5^$\circ$C, what is the minimum temperature reached by the solution (The specific heat of water = 4.18 J/g·^$\circ$C; the heat capacity of the calorimeter = 365. J/^$\circ$C.) A) 18.6^$\circ$C B) 19.7^$\circ$C C) 21.9^$\circ$C D) 27.1^$\circ$C E) 30.4^$\circ$C

Question

Quizplus · Accepted Answer

First, we need to calculate the amount of heat released by the ammonium chloride dissolving in water using the given heat of solution:

q = -nΔH
q = -(6.134 g / 53.491 g/mol) * (-15.2 kJ/mol)
q = 0.701 kJ

Next, we can use the formula for heat transfer to find the final temperature of the solution:

q = mcΔT
0.701 kJ = (65.0 g * 4.18 J/g·°C + 365. J/°C) * (Tf - 24.5°C)
Simplifying and solving for Tf:

Tf = 21.9°C

Converting to Fahrenheit:

Tf = 21.9°C * 9/5 + 32
Tf ≈ 71.4°F

Therefore, the minimum temperature reached by the solution is 21.9°C or 71.4°F, which corresponds to answer choice C.

The Heat of Solution of Ammonium Chloride Is 15 ∘\circ∘ C, What Is the Minimum Temperature Reached by the Solution

The Heat of Solution of Ammonium Chloride Is 15 $\circ$ C, What Is the Minimum Temperature Reached by the Solution