Asample of oxygen gas is collected over water at 25°c and a total pressure of 644 torr. the volume of the gas collected is 500.0 ml. what mass of oxygen is collected? the vapor pressure of water at 25°c is 23.8 torr.

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

The given parameters;

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

PV = nRT

where;

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.

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0.544g

Explanation:

To calculate the mass of oxygen collected, we can calculate the number of moles of oxygen collected and multiply this by the molar mass of the oxygen molecule.

To calculate the number of moles of oxygen collected, we can use the ideal gas equation I.e PV = nRT

Rearranging the equation, n =PV/RT

We now identify each of the terms below before substituting and calculating.

n = number of moles, which we are calculating.

R = molar gas constant = 62.64 L.Torr. K^-1. mol^-1

V = volume = 500ml : 1000ml ÷ 1L, hence , 500ml = 500/1000 = 0.5L

T = temperature = 25 degrees Celsius = 273 + 25 = 298K

P = pressure. But since the gas was collected over water, we subtract the vapour pressure of water from the total pressure = 644 - 23.8 =620.2 torr

We substitute these values into the equation to yield the following:

n = (620.2× 0.5) ÷ ( 62.64 × 298)

n = apprx 0.017 moles

To calculate the mass of oxygen collected, we need the molar mass of oxygen. The molecular formula of oxygen is O2, atomic mass of oxygen= 32g/mol

Thus the mass of oxygen collected = 0.017 mole × 32g/mol = 0.544g

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the answer is either b or c