Question

when 1.0g of ammonium nitrate, NH4NO3, dessolves in 50.0g of water, the temprature of the water drops from 25.0c to 23.32c. what is the molar heat (H) of the solution i kj/mol

Answer 1

To determine the molar heat (ΔH) of the solution in kJ/mol, we can use the formula:

ΔH = q / n

where ΔH is the molar heat of the solution in kJ/mol, q is the heat absorbed/released by the solution in Joules, and n is the number of moles of solute.

First, let's calculate the heat absorbed/released by the solution (q):

q = m × c × ΔT

where q is the heat absorbed/released in Joules, m is the mass of the solution in grams, c is the specific heat capacity of water (4.18 J/g°C), and ΔT is the change in temperature in °C.

mass of the solution = mass of ammonium nitrate + mass of water = 1.0 g + 50.0 g = 51.0 g

ΔT = final temperature - initial temperature = 23.32°C - 25.0°C = -1.68°C

Now we can calculate q:

q = 51.0 g × 4.18 J/g°C × -1.68°C = -360.2184 J

Next, let's calculate the number of moles of ammonium nitrate (n):

Using the molar mass of NH4NO3:
NH4NO3 = 1(14.007) + 4(1.007) + 1(14.007) + 3(16.00)
NH4NO3 = 28.013 + 4.028 + 14.007 + 48.00
NH4NO3 = 94.050 g/mol

n = mass of ammonium nitrate / molar mass of ammonium nitrate
n = 1.0 g / 94.050 g/mol = 0.01063 mol

Finally, let's calculate the molar heat (ΔH):

ΔH = q / n
ΔH = -360.2184 J / 0.01063 mol = -33923.71 J/mol ≈ -33.92 kJ/mol

Therefore, the molar heat of the solution is approximately -33.92 kJ/mol.