The addition of 117 g potassium fluoride to 599 ml of water (d = 1.00 g/ml) causes the temperature to rise 4.00 °c. what is the δhsoln for kf?

Answer : The for KF is -5.935 KJ/mole.

Solution : Given,

Mass of potassium fluoride = 117 g

Volume of water = 599 ml

Density of water = 1 g/ml

Molar mass of potassium fluoride = 58 g/mole

Temperature =

Specific heat capacity of water =

First we have to calculate the mass of solution.

Mass of water = Density × Volume = 1 × 599 = 599 g

Mass of solution = Mass of water + Mass of KF = 599 + 117 = 716 g

Now we have to calculate the heat absorbed by the solution.

Here, we assume that the specific heat capacity of the solution is the same as that of water.

Heat absorbed by the solution =

Heat absorbed by the solution =

Since the solution absorbed heat, the dissolution of KF must be exothermic and must have released

Now we have to calculate the moles of KF dissolved in solution.

Moles of KF =

Now we have to calculate the for KF.

Therefore, the for KF is -5.935 KJ/mole.

the answer for sodium is 0.186

the answer for rubidium is 0.251

i do not get it   at all