Question

A thermos contains 80.0g of water at 23.4 degrees C. Suppose 0.200 moles of KCl are dissolved in the water. What will be the final temperature of the solution? Assume that there is no energy transfer between the solution and the thermos, and that the specific heat is 4.184J/g*degrees C. Also, the delta H of solvation for KCl at 25 degrees C is 17.1 kJ/mol.

Answer 1

32.04°C will be the final temperature of the solution.

Step-by-step explanation:

Moles of potassium chloride = 0.200 mol

MAs sof KCl= 0.200 mol × 74.5 g/mol= 14.9 g

Enthalpy of solvation of potassium nitrate =

`\Delta H_(solv)=17.1 kJ/mol`

Energy released when 0.200 moles of KCl is dissolved in water = Q

`Q=17.1kJ/mol* 0.200 mol=3.42 kJ=3420 J`

(1 kJ = 1000 J)

Heat released on dissolving 0.200 moles of KCl is equal to heat absorbed by water = Q

Mass of solution , m= 80.0 g +14.9 g = 94.9 g

Specific heat of water = c = 4.184 J/g°C

Initial temperature of the water =
`T_1=23.4^oC`

Final temperature of the water =
`T_2=?`

`Q=m* c* (T_2-T_1)`

`3420 J=94.9g* 4.184 J/g^oC* (T_2-23.4^oC)`

`T_2=32.04^oC`

32.04°C will be the final temperature of the solution.