SolutionHard
Question
Mixture of volatile components A and B has total vapour pressure (in Torr) p = 254 - 119 xA where xA is mole fraction of A in mixture. Hence P0A and P0B are (in Torr)
Options
A.254, 119
B.119, 254
C.135, 254
D.119, 373
Solution
P = XAPA0 + XBPB0 = (PA0 - PB0)XA + PB0
So PB0 = 254
PA0 - PB0 = -119 PA0 = 135
So PB0 = 254
PA0 - PB0 = -119 PA0 = 135
Create a free account to view solution
View Solution FreeMore Solution Questions
Total vapour pressure of mixture of 1 mol of volatile component A (pAo = 100 mmHg) and 3 mol of volatile component B (pB...The concentration of pollutant in ppm (w/w), that has been measured at 450 mg per 150 kg of sample is :...The osmotic pressures of 0.010 M solutions of KI and of sucrose (C12H22O11) are 0.432 atm and 0.24 atm respectively. The...Consider separate solutions of 0.500 M C2H5OH(aq), 0.100 M Mg3(PO4)2(aq), 0.250 M KBr(aq) and 0.125 M Na3PO4(aq) at 25oC...Which of the following is a buffer solution ?...