Question

A sample of 5.20ml of diethylether (C₂H₅OC₂H₅; density=0.7134g/ml) is introduces into a 6.40L vessel that already contains a mixture of N₂ and O₂ whose partial pressures are PN₂ = 0.751atm and PO₂ =0.209atm. The temp is held at 35.0C and the diethylether totally evaporates. a) Calculate the partial pressure of the diethylether b) Calculate the total pressure in the container

Answer 1

a) The partial pressure of diethylether is 0.331 atm.

b) The total pressure in the container is 1.291 atm.

Step-by-step explanation:

To calculate the partial pressure of diethylether, we use the ideal gas law,
`\(PV = nRT\)`, where
`\(P\)` is pressure,
`\(V\)` is volume,
`\(n\)` is the number of moles,
`\(R\)` is the ideal gas constant, and
`\(T\)` is temperature in Kelvin. First, we convert the volume of diethylether from milliliters to liters by dividing 5.20 ml by 1000, yielding 0.00520 L. Next, we calculate the moles of diethylether using its density and molar mass (74.12 g/mol), finding it to be 0.070 moles.

Substituting these values into the ideal gas law, along with the given temperature of 35.0°C converted to Kelvin (308.15 K), we find the partial pressure of diethylether to be 0.331 atm.

For the total pressure in the container, we sum the partial pressures of N₂, O₂, and diethylether. The partial pressures of N₂ and O₂ are given as 0.751 atm and 0.209 atm, respectively. Adding these to the partial pressure of diethylether, we obtain a total pressure of 1.291 atm.