Answer:
105.4 kJ
Explanation:
Given that:
10 mol/h of 50 mol % NaOH is mixed with enough 6.5 mol % NaOH to produce a 10 mol % aqueous NaOH solution
Now,if the moles of 6.5 mol% of NaOH is mixed with 10 mol of 50% NaOH
We have; Total moles = (M + 10) moles
moles of NaOH = ( 10 × 0.5) + ( M × 0.065)
The mixture then tends to be the the mole of NaOH solution
= [tex]\frac{(10*0.5)(M*0.065)}{(M+10)}*100[/tex]
= [tex]\frac{800}{7} moles[/tex]
The total moles = 10 + [tex]\frac{800}{7} moles[/tex]
= [tex]\frac{70 + 800}{7}[/tex]
= [tex]\frac{870}{7} moles[/tex]
10% mole of NaOH & 90% moles of H₂O
Thus, [tex]\frac {810}{7}*0.1 moles of NaOH[/tex]
= [tex]\frac {810}{7}*0.1 moles*40 g/mole NaOH[/tex]
= [tex]\frac{3480}{7 }[/tex] g of NaOH
For H₂O
= [tex]\frac {810}{7}*0.9 moles of H_2O[/tex]
= [tex]\frac {810}{7}*0.9 moles*18g/mole H_2O[/tex]
= [tex]\frac{14094}{7}[/tex] g of H₂O
Finally, Total mass of mixture = [tex](\frac{3480}{7}+\frac{14094}{7})g[/tex]
= [tex]\frac{117574}{7}g[/tex]
Heat capacity of solution = 2.8 J/g ° C
At 25 ° C, heat removed to change the temperature from 25° C to 10° C
ΔT = ( 25 - 10 ) ° C
ΔT = 15 ° C
ΔQ = mcΔT
ΔQ = [tex]\frac{117574}{7}g[/tex] * 2.8 J/g ° C * 15° C
ΔQ = 105444 J
ΔQ = 105.4 kJ
Hence, the amount of heat to be removed to keep the final solution at 10 ° C = 105.4 kJ
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
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
