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

`63.09KJ/mol`

Step-by-step 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(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(1~KJ)/(1000~J)=32~KJ`

Finally, we can divide by the moles of LiCl:

`(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!