Question

what is the change in temperature when 3.00 grams of lithium chloride is dissolved in 100 mL of water

Answer 1

When one mole of LiCl is dissolved in water, amount of heat generated = Enthalpy of dissolution of LiCl in water = -37.03 kJ/mol

Mass of LiCl taken = 3 g

Moles of LiCl taken = Mass/MW = 3/42.4 = 0.071

So, heat released = 37.03*0.071 = 2.629 kJ = 2629 J

Assuming that density of water is equal to 1 g/mL.

Mass of water taken = Density*Volume = 1*100 = 100 g

Mass of solution = 103 g

Also assuming that heat capacity of solution = heat capacity of water = 4.18 J/(g.C)

Heat absorbed by solution = Heat released on dissolution = 2629 J

Using formula:

Q = m*C*dT

Putting values:

2629 = 103*4.18*dT

Solving we get:

dT = 6.11

So, temperature increase by approximately 6.11⁰C.