CAlculate the partial pressure of diethylether. And calculate the total pressure on the container

Question

A sample of 5.50 mL of diethylether (C2H5OC2H5; density= 0.07134 g/mL) is introduced into a 6.50 -L vessel that already contains a mixture of N2 and O2, whose partial pressures are Pn2= 0.751 atm and Po2= 0.209 atm. The temperature is held at 35 degrees C, and the diethylether totally evaporates. 

Answer 1

Assuming you mean the density of ether is 0.7134 g/ml and NOT 0.07134 g/ml ...
molar mass C2H5OC2H5 = 94.28 g/mole
mass C2H5OC2H5 = 5.50 ml x 0.7134 g/ml = 3.9237 g
moles C2H5OC2H5 = 3.9237 g x 1 mole/74.12 g = 0.05294 moles
PV = nRT where P=?; n=0.05294 moles; V=6.50L; R=0.0821 Latm/Kmol; T=35ºC +273 = 308K
P = nRT/V = (0.05294)(0.0821)(308)/6.50
P = 0.206 atm (to 3 significant figures (this is the partial pressure of diethyl ether). 

TOTAL PRESSURE = 0.206 atm + 0.751 atm + 0.209 atm = 1.17 atm (3 significant figures)