coffee cup calorimeter contained 50.0 mL of a 1.33 M KOH solution at an initial
ID: 1061728 • Letter: C
Question
coffee cup calorimeter contained 50.0 mL of a 1.33 M KOH solution at an initial temperature of 23.3 ºC. A student quickly add 40.0 mL of 1.54 M HNO3 to the calorimeter. The maximum temperature reached during the neutralization reaction was 29.1ºC.
assume specific heat of solution is the same as specific heat of water = 4.18 J/gºC
assume density of both KOH and HNO3 are the same as water = 1.00 g/mL
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This is more of a question regarding the question itself.
What's the difference in finding the mass using Mass = moles*Molar mass = (Molarity×Volume)*Molar mass
and finding mass using Mass = Density*Volume
There is obviously a big difference in these values.
For example, finding the mass of KOH using molarity and volume
Mass = (1.33)*(50*10^-3)*56.105 = 3.73g
Using density it'll be Mass = 50* 1 = 50g
What's the difference, and which method is the most correct?
Explanation / Answer
given that
50.0 mL of a 1.33 M KOH
40.0 mL of 1.54 M HNO3
initial temperature = 23.3 ºC
final temperature = 29.1ºC
specific heat of water = 4.18 J/gºC
density of both KOH and HNO3 = 1.00 g/mL
The reaction of KOH and HNO3 is:
KOH + HNO3 ----> KNO3 + H2O
Volume of KOH = 50.0 ml
Volume of HNO3 = 40.0 ml
Total volume = 50.0 + 40.0 = 90.0 ml
Density = 1.00 g/ml
Mass of 90.0 ml = 90.0 ml*1.00 g/ml = 90.0 g
m = mass = 90.0g
Specific heat of water = c = 4.18 J/g oC
Temperature rise = T = 29.1-23.3 = 5.8 oC
q = mcT = 90.0*4.18*5.8 = 2181.96J
number of mole of HNO3 = molarity * volume in L
= 1.54*0.040
= 0.0616 moles
Hrxn of HNO3 =
=2181.96 J *1.0 KJ/1000 J)/0.01616 moles
= -35.42 kJ/mol
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