Question

When a 21.5-g sample of LiCl was added to 195 g of water at a temperature of 20.00°C in a calorimeter, the temperature increased to 39.26°C. How much heat is involved in the dissolution of the LiCl?

Answer 1

Answer:

$63.09KJ/mol$

Explanation:

In this case, to calculate the heat of solution (KJ/mol) we have to take into account the mass of water, the specific heat of the water and the temperature change, so:

$m=195~g$

$c=4.18~J/g^{\circ}C$

Δ$T=39.26^{\circ}C$

With this in mind, we can use the equation:

$Q=m*c*T$

If we plug the values into the equation we will have:

$Q=195~g*4.18~J/g^{\circ}C*39.26^{\circ}C$

$Q=32000.83~J$

Now, with the mass value (21.5 g) and the molar mass of LiCl (42.39g/mol) we can calculate the moles of LiCl:

$21.5~g~LiCl\frac{1~mol~LiCl}{42.39~g~LiCl}=0.507~mol~LiCl$

Now, in the heat of solution, we have KJ/mol units. Therefore, we have to convert from J to KJ:

$32000.83~J\frac{1~KJ}{1000~J}=32~KJ$

Finally, we can divide by the moles of LiCl:

$\frac{32~KJ}{0.507~mol}=63.09KJ/mol$

So, for each mole of LiCl, we have 63.09 KJ involved in the dissolution process.

I hope it helps!