Chemistry Tutor Site - Problem
What is the equilibrium partial pressure of water vapor above a mixture of 24.0 g H2O and 48.0 g CH3CH2OH at 25 °C. The partial pressure of pure water at 25.0 °C is 23.8 mm Hg. Assume ideal behavior for the solution.
Chemistry Tutor Site - Answer
24.0 g / 18.02 g/mol = 1.33 moles water
48.0 g / 46 g/mol= 1.04 moles CH3CH2OH
total moles = 1.33 + 1.04 = 2.37
Moles fraction water = 1.33 / 2.37 = 0.561
p = p°(water ) X (water) => Raoult’s law
p = 23.8 x 0.561 = 13.4 mm Hg
October 13th, 2008
Chemistry Tutor Site-Chemistry Help
A Student has a sample that contains both KBr and KI in unknown quantities. If the Student takes the sample with a total mass of 5.00g and contains 1.51g K, what are the percentages of KBr and KI in the sample by mass?
Chemistry Tutor Site-Answer
To solve this problem we use the mass of K (Potassium) as a reference, because both KBr and KI contain K.
Mr KBr = 39.10 + 79.90 = 119 g/mol
Mr KI = 39.10 + 126.9 = 166 g/mol
We assume that the mass of K Br are x g, thus the mas of KI are (5.00-x) g and you should remember how you calculate the mass of elements in the compound using this formula:
mass of K in KBr = Ar K/Mr KBr x Mass of KBr
Using the same method we can calculate the mass of K inKI, thus the total mass of K in KBr and KI are:
Total mass of K = mass of K in KBr + mass of K in KI
1.51 g = (ArK/MrKBr).mass of KBr + (ArL/MrKI).mass of KI
1.51 g = (39.10/119).x + (39.10/166).(5.00-x) g
1.51 g = 0.329x + 0.236 (5.00-x) g
1.51 g = 0.329x + 1.18 g - 0.236x
0.33 g = 0.093x
x = 3.548 g
Because x = mass of KBr thus the mass of KBr = 3.548 g
mass of KI = (5.00-x) = 5.00 - 3.548 = 1.452 g
and the mass percentage of KBr and KI in the sample are:
%KBr = mass KBr/mass sample .100% = 3.548/5.00 . 100% = 70.96%
%KI = 100% - 70.96% = 29.04%
October 13th, 2008