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).

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

Explanation:

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

Density of solution = 1.20 g/mL

Volume of solution = 150.0 mL

Putting values in above equation, we get:

The equation used to calculate heat released or absorbed follows:

q = heat absorbed or released

m = mass of solution = 180 g

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

= change in temperature = 16.01°C

Putting values in above equation:

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

where,

q = amount of heat absorbed = 12.046 kJ

n = number of moles of solid = 3.15 moles

= enthalpy change of the reaction

Putting values in above equation, we get:

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

c. heat could easily travel through the molecules in the air and escape/enter the flask. removing the air slows down the heat transfer.