A student gets the following data for the heat of neutralization trials using 1.

Question

A student gets the following data for the heat of neutralization trials using 1.0 M NaOH and
1.0 M HCl.
Trial 1
A. Mass of calorimeter /g 43.5
B. Mass HCl and calorimeter / g 64.0
C. Mass calorimeter, HCl and NaOH / g 85.0
D1. Mass HCl / g (B - A)
D2. Mass NaOH /g (C - B)
Mass of Solution (D1 + D2)
Initial temp. HCl measured / oC (therm 1) 22.1
Initial temp. HCl corrected / C o
Initial temp. NaOH measured / oC (therm 2) 21.6
Initial temp. NaOH corrected / C o
E. Average Initial Temp corrected /BC
Final temp. measured / oC (therm 1) 28.8
F. Final temp. corrected / C o
G. Temp. change corrected / C o
(F - E)

Using this data, determine the Molar Enthalpy of Neutralization for this trial.
The mcalib is 0.969 and bcalib is 2.0 C for both Thermometer 1 and Thermometer 2 and B
the calculated heat capacity of the calorimeter was 33 J °C-1

Explanation / Answer

Mass of Cold Water = 94.2-43.5 = 50.7grams

T1 = 21.3 c

T2 = 45.2 C

Tf = 30

mass of hot water = 144.7-94.2 = 50.5 grams

heat lost by hot water = heat gained by cold water + heatcapacity of calorimeter*DT

(50.5*4.18*(45.2-30)) = (50.7*4.18*(30-21.3))+(x*(30-21.3))

x = heatcapacity of calorimeter = 0.156 kj/C

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