M1V1 = M2 V2
50.0 x 0.805 = M2 x 23.55
M2 = 1.71 M
initial concentration of sodium hydroxide = 1.71 M
final concentration of sodium ions = 1.71 mol /L
total volume = 50 + 23.55 = 73.55 mL
density = 1 g / mL
mass = density x volume
= ( 1 g / mL ) x 73.55 mL
= 73.55 g
dT = 29.4 - 22.1 = 7.3 oC
Q = m Cp dT
Q = 73.55 x 4.184 x 7.3
Q = 2246.45 J
Q = 2.25 kJ
heat is liberated by this neutralization reaction = 2.25 kJ
moles of acid = 50 x 0.805 / 1000 = 0.04025
heat of neutralization = - Q / n
= -2.25 kJ / 0.04025
= -55.9 kJ / mole
heat of neutralization = -55.9 kJ / mol
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