Question

A vessel containing 18.8 g of SF4(g) and 2.17 g of CO(g) has a total pressure of 0.823 atm. What are the partial pressures of SF4 and CO?

Answer 1

Molar mass of SF4 = 1*MM(S) + 4*MM(F)

= 1*32.07 + 4*19.0

= 108.07 g/mol

Molar mass of CO = 1*MM(C) + 1*MM(O)

= 1*12.01 + 1*16.0

= 28.01 g/mol

n(SF4) = mass of SF4/molar mass of SF4

= 18.8/108.07

= 0.174

n(CO) = mass of CO/molar mass of CO

= 2.17/28.01

= 0.0775

n(SF4),n1 = 0.174 mol

n(CO),n2 = 0.0775 mol

Total number of mol = n1+n2

= 0.174 + 0.0775

= 0.2514 mol

Partial pressure of each components are

p(SF4),p1 = (n1*Ptotal)/total mol

= (0.174 * 0.823)/0.2514

= 0.569 atm

p(CO),p2 = (n2*Ptotal)/total mol

= (0.0775 * 0.823)/0.2514

= 0.254 atm

partial pressure of SF4 = 0.569 atm

partial pressure of CO = 0.254 atm