Question

A sample 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.

Answer 1

Step-by-step 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^(o)C` = (35 + 273) K = 308 K

As, Density =
`(mass)/(volume)`

0.7134 g/ml =
`(mass)/(5.10 ml)`

mass = 3.63 g

Also, No. of moles =
`\frac{mass}{\text{molar mass}}`

Hence, moles of diethyl ether will be as follows.

No. of moles =
`\frac{mass}{\text{molar mass}}`

=
`(3.63 g)/(74.12 g/mol)`

= 0.0489 mol

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

PV = nRT

`P * 6 L = 0.0489 * 0.0821 L atm/mol K * 308 K`

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.

`P_(total) = P_{N_(2)} + P_{O_(2)} + P_{C_(2)H_(5)OC_(2)H_(5)}`

= 0.752 atm + 0.206 atm + 0.206 atm

= 1.164 atm

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

Answer 2

A) partial pressure of diethylether = 0.206 atm

B) Total pressure = 1.164 atm

Step-by-step 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