When 1.0 g of ammonium nitrate, NH4NO3, dissolves in 50.0 g of water, the temperature of the water drops from 25.0C to 23.32C. What is the molar heat (H) of solution in kJ/mol?

1 answer

To calculate the molar heat of solution, we can use the equation:

H = q / n

Where:
H is the molar heat of solution (in kJ/mol),
q is the heat transferred (in kJ),
and n is the number of moles of solute.

First, let's calculate the heat transferred (q).
The heat transferred can be calculated using the equation:

q = m * c * ΔT

Where:
q is the heat transferred (in kJ),
m is the mass of the solvent (in g),
c is the specific heat capacity of water (4.18 J/g°C),
ΔT is the change in temperature (in °C).

Given:
Mass of water (solvent), m = 50.0 g
Initial temperature, T1 = 25.0°C
Final temperature, T2 = 23.32°C

ΔT = T2 - T1 = 23.32°C - 25.0°C = -1.68°C

q = 50.0 g * 4.18 J/g°C * (-1.68°C)

Convert grams to kilograms:
q = 0.050 kg * 4.18 J/g°C * (-1.68°C)

Now we have q in Joules. To convert it to kilojoules:
q = 0.050 kg * 4.18 J/g°C * (-1.68°C) /1000 = -0.0013902 kJ

Next, let's calculate the number of moles of NH4NO3.

Given:
Mass of ammonium nitrate, m = 1.0 g
Molar mass of ammonium nitrate, MM = (1 * 14) + (4 * 1) + (1 * 14) + (3 * 16) = 80 g/mol

n (in moles) = m / MM
n = 1.0 g / 80 g/mol

Now, let's calculate H.

H = q / n
H = -0.0013902 kJ / (1.0 g / 80 g/mol)

H = -0.0013902 kJ * (80 g/mol / 1.0 g)

H = -0.111216 kJ/mol

Therefore, the molar heat (H) of solution of ammonium nitrate is approximately -0.111 kJ/mol.