Question

A sample of 5.30 mLmL of diethylether (C2H5OC2H5;density=0.7134g/mL)(C2H5OC2H5;density=0.7134g/mL) is introduced into a 6.50 L L vessel that already contains a mixture of N2N2 and O2O2, whose partial pressures are PN2=0.750 atmPN2=0.750 atm and PO2=0.207 atmPO2=0.207 atm. The temperature is held at 35.0 ∘C∘C, and the diethylether totally evaporates

Answer 1

1.16atm

Step-by-step explanation:

We are going to derive the mass of ether from density

mass=density *volume

Also moles=mass/molecular mass

molar mass C2H5OC2H5 =74.12 g/mole

the density of ether is 0.7134 g/ml

mass C2H5OC2H5 = 5.30 ml x 0.7134 g/ml = 3.78 g

moles C2H5OC2H5 =3.78 g x 1 mole/74.12 g = 0.0509 moles

PV = nRT where P=?; n=0.0509 moles; V=6.50L; R=0.0821 Latm/Kmol; T=35ºC +273 = 308K

P = nRT/V = 0.0509)(0.0821)(308)/6.50

P = 0.198 atm (to 3 significant figures (this is the partial pressure of diethyl ether).

TOTAL PRESSURE

P1+p2+p3

= 0.198 atm + 0.750 atm + 0.207 atm =1.1550atm

1.16atm(3 significant figures)