A 6.3 g sample of solid NH4NO3 is dissolved in 50g of water, causing the temperature to drop from 24°C to 19°C. Calculate the ∆Hsolution in J/mol.
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To calculate the ∆Hsolution for the dissolution of NH4NO3, we need to use the equation:
∆Hsolution = q / n
where q is the heat absorbed or released during the dissolution process, and n is the number of moles of NH4NO3 that dissolved.
To find the value of q, we can use the equation:
q = m × c × ∆T
where m is the mass of the water, c is the specific heat capacity of water, and ∆T is the change in temperature.
First, let's calculate the values we need:
m (mass of water) = 50 g
c (specific heat capacity of water) = 4.18 J/g°C (rounded value, at 25°C)
∆T (change in temperature) = 19°C - 24°C = -5°C
Now, let's calculate q:
q = 50 g × 4.18 J/g°C × (-5°C)
q = -1045 J
Next, we need to find the number of moles of NH4NO3 that dissolved. To do this, we use the equation:
n = m / M
where m is the mass of NH4NO3, and M is the molar mass of NH4NO3.
m (mass of NH4NO3) = 6.3 g
M (molar mass of NH4NO3) = 80 g/mol (you can calculate this by adding up the atomic masses of the elements in NH4NO3: 14.01 + 1.01 + 14.01 + 14.01 + 16.00 + 16.00 + 16.00 = 80.04 g/mol)
Now, let's calculate n:
n = 6.3 g / 80 g/mol
n = 0.079 mol
Finally, we can calculate the ∆Hsolution:
∆Hsolution = q / n
∆Hsolution = -1045 J / 0.079 mol
∆Hsolution = -13227.84 J/mol
Therefore, the ∆Hsolution for the dissolution of NH4NO3 is -13227.84 J/mol.