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?

Answer 1

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

H_rxn of HNO3 =

=2181.96 J *1.0 KJ/1000 J)/0.01616 moles

= -35.42 kJ/mol