Answer to Question #68484 in General Chemistry for Johnny

Question

The molar heat of a solution, ʌH soln , of NaOH is -44.51 kJ/mol . In a certain experiment, 36.5 grams of NaOH is completely dissolved in 100 mL of water at 22.0° C in a foam cup calorimeter. Calculate the final temperature of the water.
Please explain this to me.

Answer 1

The enthalpy change of solution is the enthalpy change when 1 mole of an ionic substance dissolves in sufficient water to give a solution of infinite dilution.
In our task we have 36.5 g NaOH and we should calculate the quantity in moles:
n(NaOH)= m (NaOH)M(NaOH)= 36.540=0.9125 (mol)
We know that 44.51 kJ are evolved from 1 mol NaOH, but we have 0.9125 mol. So the calculation of heat will be next:
Q=-∆H1=∆H×n(NaOH)1 mol= -44.51×0.91251=40.62 (kJ)
From equitation we know that:
Q=c×m×(T2-T1)⇒T2=Qc×m+T1
c – heat capacity of water, m – mass of water
T2=40.62*10004.1813×100+(273+22)=392.1 K= 119.1℃
Answer: 119.1°C.