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
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.
To solve for the molar heat (H) of the solution in kJ/mol, you can use the equation:
H = (mass of water) x (change in temperature) / (moles of solute)
First, let's find the moles of ammonium nitrate (NH4NO3) in the solution:
Molar mass of NH4NO3 = (1 x 1.01) + (4 x 1.01) + 1.008 + (3 x 14.01) = 80.04 g/mol
moles of NH4NO3 = mass / molar mass = 1.0 g / 80.04 g/mol = 0.01249 mol
Now we can calculate the molar heat (H):
H = (mass of water) x (change in temperature) / (moles of solute)
= 50.0 g x (23.32°C - 25.0°C) / 0.01249 mol
= 50.0 g x (-1.68°C) / 0.01249 mol
= -84 kJ/mol
Therefore, the molar heat of the solution is -84 kJ/mol. The negative sign indicates that the reaction is exothermic, meaning it releases heat to the surroundings.
To find the molar heat (H) of the solution in kJ/mol, you can use the equation:
H = q / n
Where:
H is the molar heat of the solution in kJ/mol
q is the heat exchanged in the reaction
n is the number of moles of ammonium nitrate (NH4NO3)
First, we need to calculate the heat exchanged in the reaction using the formula:
q = mcΔT
Where:
q is the heat exchanged in the reaction
m is the mass of the water in grams
c is the specific heat capacity of water (about 4.18 J/g°C or 1.00 cal/g°C)
ΔT is the change in temperature of the water (initial temperature - final temperature)
Given:
Mass of water (m) = 50.0 g
Specific heat capacity of water (c) = 4.18 J/g°C (or 1.00 cal/g°C)
Initial temperature (T1) = 25.0°C
Final temperature (T2) = 23.32°C
First, convert the units of mass and temperature to the SI units:
Mass of water (m) = 50.0 g = 0.050 kg
Now, calculate the heat exchanged (q):
q = mcΔT
= (0.050 kg) * (4.18 J/g°C) * (23.32°C - 25.0°C)
Next, to calculate the number of moles of ammonium nitrate (NH4NO3), we use the formula:
n = m / M
Where:
n is the number of moles of NH4NO3
m is the mass of NH4NO3 in grams
M is the molar mass of NH4NO3
Given:
Mass of NH4NO3 (m) = 1.0 g
Molar mass of NH4NO3 (M) = 80.04 g/mol (14.01 g/mol for N, 1.01 g/mol for H, and 16.00 g/mol for O)
Now, calculate the number of moles (n):
n = m / M
= 1.0 g / 80.04 g/mol
Finally, substitute the calculated values of q and n into the equation for molar heat (H):
H = q / n
Solve for H to find the molar heat of the solution in kJ/mol.
Note: Make sure to perform the calculations using the correct units to obtain the final answer in kJ/mol.