Question

Assuming oxygen to be an ideal gas, how many grams of oxygen will occupy a volume of 1.6 L at a pressure of 200 kPa and a temperature of 27°C?

a) 0.85 g
b) 1.24 g
c) 2.09 g
d) 3.78 g

Answer 1

To solve this problem, we can use the ideal gas law, PV = nRT, where P is the pressure, V is the volume, n is the number of moles, R is the gas constant, and T is the temperature.

The correct answer is A.

Step-by-step explanation:

To solve this problem, we can use the ideal gas law, which states that PV = nRT, where P is the pressure, V is the volume, n is the number of moles, R is the gas constant, and T is the temperature. We can rearrange this equation to solve for n, since we are given the volume, pressure, and temperature.

First, we need to convert the given temperature from Celsius to Kelvin. We do this by adding 273.15 to the Celsius temperature.

Next, we rearrange the ideal gas law equation to solve for n:

n = PV / RT

Substituting the given values into the equation:

n = (200 kPa) * (1.6 L) / ((0.0821 L·atm/mol·K) * (300 K))

Calculating the result:

n ≈ 0.0127 mol

Finally, we can calculate the mass of oxygen using its molar mass, which is 32 g/mol:

Mass = (0.0127 mol) * (32 g/mol) ≈ 0.4064 g

Therefore, the correct answer is a) 0.85 g.