When 0.320 moles of NH4Cl is dissolved in 1.00 L of water, the temperature drops by 13.7 degrees C. What is the heat of solution of ammonium chloride? The solution has a density of 1.0012 g/mL and a specific heat of 0.9818 cal/g degrees C.
q = mass water x specific heat water x delta T.
Use density to determine mass water.
delta H soln = q/0.320 for cal/mol
can you be more specific? that equation does not make any sense.
To calculate the heat of solution, we need to use the equation:
Heat of solution = mass of solution x specific heat x change in temperature
First, we need to find the mass of the solution. We can calculate it using the density of the solution and the volume of the solution:
mass of solution = density x volume
Given that the density of the solution is 1.0012 g/mL and the volume is 1.00 L, we can convert the volume to mL:
volume = 1.00 L x 1000 mL/L = 1000 mL
Now we can calculate the mass of the solution:
mass of solution = 1.0012 g/mL x 1000 mL = 1001.2 g
Next, we need to calculate the change in temperature. Given that it drops by 13.7 degrees C, the change in temperature is -13.7 degrees C.
Now we can plug the values into the equation:
Heat of solution = 1001.2 g x 0.9818 cal/g degrees C x (-13.7 degrees C)
Calculating the expression:
Heat of solution ≈ -14258 cal
Therefore, the heat of solution of ammonium chloride is approximately -14258 cal.