A 900 L tank at 856 C is filled with 693 g of carbon monoxid
Solution
Volume = V= 9.00L
T= 8.56C= 8.56+273= 281.56K
R= 0.0821 L-atm/mol-K
mass of CO= 6.93 grams
Molar mass of CO= 12.0+15.999= 27.999 gram/mole
number of moles of CO= mass/molar mass = 6.93/27.999 = 0.248 moles
number of moles of CO= 0.248 moles
mass of BF3 = 14.3 grams
molar mass of BF3 = 10.811+3x18.998= 67.805 gram/mole
number of moles of BF3= 14.3/67.805 = 0.211 moles
Total number of moles = 0.248 + 0.211 = 0.459 moles
n-=0.459 moles
PV=nRT
P= nRT/V = 0.459 x0.0821x 281.56/9.00 = 1.179 atm
P= 1.179 atm
mole fraction of CO = number of moles of CO/total number of moles
molefraction of CO= 0.248/0.459= 0.54
molefraction of BF3 = 0.211/0.459 =0.46
partial pressure of CO= total pressure x molefraction of CO
partial pressure of CO= 1.179 x0.54 = 0.637 atm
partial pressure of BF3 = 1.179 x0.46= 0.542 atm
Total pressure = 1.179 atm

