Question

To a calorimeter, which contained 80 g water at 25.0ºC was added 4.0 g NaOH. After the NaOH had dissolved the temperature was 34.0ºC. From this information calculate the enthalpy (ΔH, kJ / mol) of the dissolution of NaOH in water.

Answer 1

Water(H₂O)=80g at 25$\degree C$

NaOH=4g

Solution's Temp=34$\degree C$

Enthalpy =mc$\Delta$ T

Assume the Calorimeter does not absorb heat

Assume density of solution is same as that of water

Assume that the specific heat capacity of the solution is same as that of water

Therefore:

$Q=mc\Delta T$

But m=(4+80)g=84g

$\Delta T=(34-25)\degree C$ =9$\degree C$

$Q=84g×4.18J/g/\degree C ×9\degree C$

$Q=3160.1J$

But 1J =0.001kJ

$\therefore3160.1J\to$

$3160.1J ×0.001kJ$

$Q=3.1601kJ$

Calculate the number of moles of NaOH in the solution

$Moles=\frac{mass}{molar mass}$

=$\frac{4}{40}=0.1moles$

If 0.1moles=3.1601kJ

$\therefore 1mole\to$

$\frac{1mole ×3.1601kJ}{0.1moles}$

=$31.601kJ/mol$