Asample of 5.10 ml of diethylether (c2h5oc2h5; density = 0.7134 g/ml) is introduced into a 6.00 -l vessel that already contains a mixture of n2 and o2, whose partial pressures are pn2 = 0.752 atm and po2 = 0.206 atm. the temperature is held at 35.0 °c, and the diethylether totally evaporates.
a) calculate the partial pressure of the diethylether.
b) calculate the total pressure in the container.

A) partial pressure of diethylether = 0.206 atm

B) Total pressure =  1.164 atm

Explanation:

Given: Density of diethylether = 0.7134 g/mL, Volume of diethylether= 5.10 mL, Total volume = 6 L

Molar mass of diethylether = 74.12 g/mol

A) As we know, Density = given mass ÷ volume

⇒ given mass of diethylether = density × volume =  0.7134 g/mL ×5.10 mL = 3.64 g

As the number of moles of diethylether = given mass  ÷ molar mass = 3.64 g   ÷ 74.12 g/mol = 0.049 mol

To find out the pressure of diethylether, we use the ideal gas equation:

PV = nRT

Here, P is the partial pressure of the gas, n is the number of moles of gas = 0.049 mol, V is the total volume = 6 L, R is the gas constant = 0.08206 L·atm/(mol·K) and T is the temperature = 35°C = 35 + 273 = 308 K      (0 °C = 273 K)

⇒ P = nRT ÷ V  = (0.049 mol × 0.08206 L·atm/(mol·K) × 308 K) ÷ 6 L

⇒ P = 0.206 atm  = partial pressure of diethylether : pC₂H₅OC₂H₅

B) Given: partial pressure of: pN₂= 0.752 atm, pO₂ = 0.206 atm and pC₂H₅OC₂H₅ = 0.206 atm

As the Total pressure of all the gases = sum of partial pressure of the gases

⇒ Total pressure = pN2 + pO₂ + pC₂H₅OC₂H₅ = 0.752 atm +  0.206 atm 0.206 atm = 1.164 atm

Explanation:

(A)   The given data is as follows.

Volume of diethyl ether = 5.10 ml

Density of diethyl ether = 0.7134 g/ml

Partial pressure of nitrogen = 0.752 atm

Partial pressure of oxygen = 0.206 atm

Temperature = = (35 + 273) K = 308 K

As,       Density =  

           0.7134 g/ml =

                 mass = 3.63 g

Also,    No. of moles =

Hence, moles of diethyl ether will be as follows.

           No. of moles =

                                 =    

                                 = 0.0489 mol

Hence, using the ideal gas equation we will find the partial pressure of diethyl ether as follows.

                   PV = nRT

                 P = 0.206 atm

Hence, the partial pressure of diethyl ether is 0.206 atm.

(B)  As it is known that total pressure of a mixture of gases is equal to the sum of partial pressures of all the gases present in the mixture.

Therefore, we will calculate the total pressure as follows.

                     = 0.752 atm + 0.206 atm + 0.206 atm

                     = 1.164 atm

Hence, the total pressure in the container is 1.164 atm.

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