The heat of solution of ammonium nitrate is 26.2 kJ/mol. If a 5.368 g sample of NH₄NO₃ is added to 40.0 mL of water in a calorimeter at 23.5^$\circ$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 = 650. J/^$\circ$C.) A) -7.7^$\circ$C B) 14.3^$\circ$C C) 20.8^$\circ$C D) 21.4^$\circ$C E) 25.6^$\circ$C

Question

Quizplus · Accepted Answer

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

q = (26.2 kJ/mol) x (5.368 g / 80.04 g/mol) = 1.76 kJ

Next, we need to calculate the amount of heat absorbed by the water and the calorimeter:

q = (m_water x c_water x ΔT) + (C_calorimeter x ΔT)

where ΔT is the change in temperature and m_water is the mass of water used. We can rearrange this equation to solve for ΔT:

ΔT = q / (m_water x c_water + C_calorimeter)

ΔT = (1.76 kJ) / [(40.0 g) x (4.18 J/g·°C) + (650. J/°C)]

ΔT = -21.4°C

Finally, we need to convert the temperature change from Celsius to Fahrenheit:

ΔT = -21.4°C x (9/5) = -38.5°F

To find the minimum temperature reached by the solution, we need to subtract the temperature change from the initial temperature of 23.5°C (or 74.3°F):

T_minimum = 23.5°C - 21.4°C = 2.1°C

T_minimum = 74.3°F - 38.5°F = 35.8°F

Therefore, the best choice is D with the minimum temperature reached by the solution of 21.4F1F1F1S1F1F1F10C or 38.5°F.

The Heat of Solution of Ammonium Nitrate Is 26 ∘\circ∘ C, What Is the Minimum Temperature Reached by the Solution

The Heat of Solution of Ammonium Nitrate Is 26 $\circ$ C, What Is the Minimum Temperature Reached by the Solution