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?

Respuesta :

Answer:

[tex]63.09KJ/mol[/tex]

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:

[tex]m=195~g[/tex]

[tex]c=4.18~J/g^{\circ}C[/tex]

Δ[tex]T=39.26^{\circ}C[/tex]

With this in mind, we can use the equation:

[tex]Q=m*c*T[/tex]

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

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

[tex]Q=32000.83~J[/tex]

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

[tex]21.5~g~LiCl\frac{1~mol~LiCl}{42.39~g~LiCl}=0.507~mol~LiCl[/tex]

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

[tex]32000.83~J\frac{1~KJ}{1000~J}=32~KJ[/tex]

Finally, we can divide by the moles of LiCl:

[tex]\frac{32~KJ}{0.507~mol}=63.09KJ/mol[/tex]

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

I hope it helps!

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