A 250. mL sample of oxygen is collected over water at 25 °C and 760.0 torr pressure. What is the partial pressure
of the oxygen? (Vapor pressure of water at 25°C is 23.8 mm Hg)

Respuesta :

Answer:The mass of oxygen gas sample collected over water is 0.544 g.

The given parameters;

temperature of the oxygen gas, t = 25 °C

total pressure of the gas, P = 644 torr

volume of the gas, V = 500 mL

vapor pressure = 23.8 torr

molar mass of oxygen gas = 32 g/mol

The number of moles of the gas collected is calculated by using ideal gas equation as follows;

PV = nRT

where;

R is the ideal gas constant = 62.36 L.torr/mol.K

P is the pressure of the gas = 644 torr - 23.8 torr = 620.2 torr

T is the temperature of the gas = 25 + 273 = 298 K

V is the volume of the gas = 500 mL = 0.5 L

The number of moles of the  gas is calculated as follows;

The mass of oxygen gas collected is calculated as follows;

m = 0.017 (32 g/mol)

m = 0.544 g

Thus, the mass of oxygen gas sample collected over water is 0.544 g.

Explanation:

The partial pressure of oxygen is; 760.0 torr - 23.8  torr = 736.2 torr.

What is partial pressure?

We know that in a mixture of gases, the pressure that is exerted by each gas is called the partial pressure of the mixture.

In this case, we have the total pressure of the mixture as 760.0 torr and the vapor pressure of water as 23.8 mm Hg or 23.8  torr) at that temperature. Hence, the partial pressure of oxygen is; 760.0 torr - 23.8  torr = 736.2 torr.

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