Question

In the laboratory a general chemistry student finds that when 1.75 g of CoSO₄(s) are dissolved in 107.60 g of water, the temperature of the solution increases from 23.74 to 25.76 °C. 

The heat capacity of the calorimeter (sometimes referred to as the calorimeter constant) was determined in a separate experiment to be 1.74 J/°C.

Based on the student's observation, calculate the enthalpy of dissolution of CoSO₄(s) in kJ/mol. 

Assume the specific heat of the solution is equal to the specific heat of water.

Answer 1

Mass of the solution is $(9.26+106)g=115.26g$

Specific heat capacity of the solution is equal to specific heat capacity of the water.

C= 4.186J/(g.°C)

Change in $Temp = T_f-T_i$

T_f=20.92°C

T_i=23.80°C

Change in $Temp =20.92-23.80=-2.88°C$

$Q_{solution}=mC×change in Temp$

$Q_{solution} =-115.26×4.186×2.88J$

$Q_{solution} =-1389.538J$

$Q_{reaction}=-( Q_{solution}+Q_{calorimeter})$

$Q_{ calorimeter}=z×change in Temp$

$z=heat capacity of calorimeter=1.74J/°C$

$Q_{calorimeter}=-(1.74×2.88)J$

$Q_{calorimeter}=-5.011J$

$Q_{reaction}=-(Q_{calorimeter}+Q_{solution})$

$Q_{reaction}=-(-1389.538+(-5.011))J$

$Q_{reaction}=1394.549J$

$ΔH_{dissolution}=Q_{reaction}/n$

$n = number of moles =9.26/212.81=0.044moles$

$ΔH_{dissolution}=(1394.549/0.044)J/mol$

$ΔH_{dissolution}=31,694.2955J/mol=31.694KJ/mol$