Question

3.15 mol of an unknown solid is placed into enough water to make 150.0 mL of solution. The solution's temperature increases by 16.01°C. Calculate ∆H for the dissolution of the unknown solid. (The specific heat of the solution is 4.18 J/g･°C and the density of the solution is 1.20 g/mL).

Answer 1

Answer: The enthalpy change of the unknown solid is 3.824 kJ/mol

Explanation:

To calculate the mass of solution , we use the equation:

$\text{Density of substance}=\frac{\text{Mass of substance}}{\text{Volume of substance}}$

Density of solution = 1.20 g/mL

Volume of solution = 150.0 mL

Putting values in above equation, we get:

$1.20g/mL=\frac{\text{Mass of solution}}{150.0mL}\\\\\text{Mass of solution}=(1.20g/mL\times 150.0mL)=180g$

The equation used to calculate heat released or absorbed follows:

$Q=m\times c\times \Delta T$

q = heat absorbed or released

m = mass of solution = 180 g

c =  specific heat capacity = 4.18 J/g°C

$\Delta T$ = change in temperature = 16.01°C

Putting values in above equation:

$Q=180\times 4.18\times 16.01=12045.9J=12.046kJ$

To calculate the enthalpy change of the reaction, we use the equation

$\Delta H_{rxn}=\frac{q}{n}$

where,

q = amount of heat absorbed = 12.046 kJ

n = number of moles of solid = 3.15 moles

$\Delta H_{rxn}$ = enthalpy change of the reaction

Putting values in above equation, we get:

$\Delta H_{rxn}=\frac{12.046kJ}{3.15mol}=3.824kJ/mol$

Hence, the enthalpy change of the unknown solid is 3.824 kJ/mol