Answer:
option 4, none of the above
Question (in proper order):
10 mol/h of 50 mol % NaOH is mixed with enough 6.5 mol % NaOH to produce a 10 mol % aqueous NaOH solution. The NaOH and water are initially at 25 °C. Assume the heat capacity of solution is 2.8 J/g C. How much of heat to be removed to keep the final solution at 10°C?
1. 600 kJ
2. 700 kJ
3. 785 kJ
4. None of above
5. 550 kJ
Step-by-step explanation:
let the no of moles of 6.5% NaOH = M
10 moles of 50% NaOH + M moles of 6.5% NaOH = [10 + M] moles of 10% NaOH
this can be interpreted thus:
10 x 50% + M x 6.5% = [10 + M] x 10%
(10 x 0.5) + (M x 0.065) = [10 + M] x 0.1
5 + 0.065M = 1 + 0.1M
5 - 1 = 0.1M - 0.065M
4 = 0.035M
4/0.035 = M
M = 114.29 moles
hence, total moles of the resulting aqueous (solution containing water) solution = 10 + M
= 10 + 114.29
= 124.29 moles
note that, if the solution is 10% NaOH, then it contains (100 - 10)% water
that is 90% water
interpreted thus
10% of 124.29 moles of NaOH + 90% of 124.29 moles H₂O
0.1 x 124.29 of NaOH + 0.9 x 124.29 of H₂O
12.429 moles of NaOH + 111.861 moles of H₂O
Total Mass of the Solution is calculated thus
mass of NaOH = (23 + 16 + 1) = 40g
mass of H₂O = (2 x 1 + 16) = 18g
12.429 moles x 40g/mole + 111.861 moles x 18g/mole
497.16g + 2013.498g
2510.658g is the total weight of the solution
from the question,
the specific heat capacity of the solution c = 2.8J/g.C
To calculate the heat change of the solution
we use the formula
ΔQ = mcΔT
where ΔQ = heat change or heat removed
m = mass of the solution, 2510.658g
c = 2.8J/g C
ΔT = Ф2 - Ф1
Ф2 = 10 °C
Ф1 = 25 °C
ΔT = 10 - 25 = -15 °C
ΔQ = 2510.658g x 2.8J/g °C x -15 °C
= - 105447.636Joules
= - 105.45kJ the minus sign implies that the heat is removed from the solution
= since the answer is not in the options, then option 4 is the right answer