Respuesta :
Answer:- [tex]\Delta H_s_o_l_n=-37.2\frac{kJ}{mol}[/tex]
Solution:- First of all we calculate the heat absorbed or released when the solute is added to the solvent. Here the solute is LiCl and the solvent is water.
To calculate the heat absorbed or released we use the formula:
[tex]q=ms\Delta T[/tex]
q = heat absorbed or released
m = mass of solution
s = specific heat capacity
and [tex]\Delta T[/tex] = change in temperature
mass of solution = mass of solute + mass of solvent
mass of solution = 5.00 g + 100.0 g = 105.0 g
(note:- density of pure water is 1 g per mL so the mass is same as its volume)
[tex]\Delta T[/tex] = 33.0 - 23.0 = 10.0 degree C
s = [tex]4.18\frac{J}{g.^0C}[/tex]
Let's plug in the values in the formula and calculate q.
q = [tex]105.0g(4.18\frac{J}{g.^0C})(10.0^0C)[/tex]
q = 4389 J
To calculate the enthalpy of solution that is [tex]\Delta H_s_o_l_n[/tex] we convert q to kJ and divide by the moles of solute.
moles of solute = [tex]5.00g(\frac{1mol}{42.39g})[/tex]
= 0.118 moles
q = [tex]4389J(\frac{1kJ}{1000J})[/tex] = 4.389 kJ
[tex]\Delta H_s_o_l_n=\frac{4.389kJ}{0.118mol}[/tex]
[tex]\Delta H_s_o_l_n[/tex] = [tex]37.2\frac{kJ}{mol}[/tex]
Since the heat is released which is also clear from the rise in temperature of the solution, the sign of enthapy of solution will be negative.
So, [tex]\Delta H_s_o_l_n=-37.2\frac{kJ}{mol}[/tex]
