Question

Student conducts an experiment to determine the enthalpy of solution for lithium chloride dissolved in water. the student combines 5.00 grams of lithium chloride with 100.0 ml of distilled water. the initial temperature of the water is 23.0°c and the highest temperature after mixing reaches 33.0°c. assume a density of 1.00 g/ml and a specific heat of 4.18 .

Answer 1

Answer:-  $\Delta H_s_o_l_n=-37.2\frac{kJ}{mol}$

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:

$q=ms\Delta T$

q = heat absorbed or released

m = mass of solution

s = specific heat capacity

and $\Delta T$ = 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)

$\Delta T$ = 33.0 - 23.0 = 10.0 degree C

s = $4.18\frac{J}{g.^0C}$

Let's plug in the values in the formula and calculate q.

q = $105.0g(4.18\frac{J}{g.^0C})(10.0^0C)$

q = 4389 J

To calculate the enthalpy of solution that is $\Delta H_s_o_l_n$ we convert q to kJ and divide by the moles of solute.

moles of solute = $5.00g(\frac{1mol}{42.39g})$

= 0.118 moles

q = $4389J(\frac{1kJ}{1000J})$ = 4.389 kJ

$\Delta H_s_o_l_n=\frac{4.389kJ}{0.118mol}$

$\Delta H_s_o_l_n$ = $37.2\frac{kJ}{mol}$

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,  $\Delta H_s_o_l_n=-37.2\frac{kJ}{mol}$