Please help with full answers. The enthalpy of solution of KBr in water is about
ID: 1023950 • Letter: P
Question
Please help with full answers.
The enthalpy of solution of KBr in water is about +198 kJ/mol (endothermic). Nevertheless, the solubility of KBr is relatively high. Explain why. A sample consisting of various gases contains 3.5*10^-6 mole fraction of radon. This gas at a total pressure of 36 atm is shaken with water at 30 degree C. Assume that the solubility of radon in water with 1 atm pressure of the gas over the solution at 30 degree C is 7.27*10^-3M. Calculate the molar concentration of radon in the water. Caffeine (C_8H_10N_4O_2) is a stimulant found in coffee and tea. A solution of caffeine in the solvent chloroform (CHCI_3) has a concentration of 0.0520m. a) Calculate the percentage of caffeine by mass in the solution. b) Calculate the mole fraction of caffeine in the solution.Explanation / Answer
Problem 1- The enthalpy of solution of KBr in water is about +198 kJ/mol
A process may be spontaneous or not depends on its free energy change.
G is determined by both the enthalpy change and the entropy change.
The enthalpy of KBr in water very high
G = H - T S
Since Delta H is very large and positive, -T Delta S is a large negative number, and more negative than 198 kJ/mol, making Delta G negative, and the process spontaneous. So solubility of KBr in water is relatively high.
Problem 2.
Given- mole fraction of radon =3.5×106
total pressure = 36 atm
Solubility of radon at1atm = 7.27x10^-3M
Pradon = mol fraction of radon x (total p)
= 3.5x10^-6(36atm)=.000126atm
Assuming that the solubility of radon in water with 1 atm pressure of the gas over the water at 30 degrees Celsius is 7.27x10^-3 M
The Henry's law constant (k) for radon in water at this temperature
K =P/C
K = 1atm/7.27x10^-3M
k=137.55atm/M
Since temperature is at a constant and is unused in the equation
P=kC
0.000126atm = 137.55atm/M x C
C = 0.000126atm /137.55atm/M
C = 9.16 x 10^-7M
.0001023atm=(137.55atm/M)(C) ---P=kC >divide both sides by 137.55 atm/M
C =7.44x10^-7M
Problem3
Convert the .052mol caffine to grams using the gram formula wgt.
Caffeine has a wgt of 194 g/mol.
0.052 mol x 194 g/mol = 10.088 g.
Because the concentration was given in molality we know that the solvent is equal to 1 kilogram.
Using 1 kg = 1000g and that chloroform has a gram molecular wgt of 122.5 g/mol we convert 1000 g to 8.16 mol.
a) % mass of caffeine = mass of caffeine/ total mass of the solution. = 10.088/(1000+ 10.088) x 100% = 0.99 %
b) mole fraction of caffeine = moles solute /moles solution = 0.052/(0.052+8.16) = 0.0063