Calculate The volume occupied at s.t.p by 0.24g of ozone ( o=16, G.M.V=22.4dm3
n = mols = grams/molar mass = ?
Then use pV = nRT.
OR you can use the fact that 1 mol of a gas at STP occupies 22.4 L.
To calculate the volume occupied by a given mass of a gas at STP (Standard Temperature and Pressure), we can use the formula:
Volume = (mass of gas * molar mass) / (GMV)
Where:
- mass of gas is the given mass of the gas (in grams)
- molar mass is the molar mass of the gas (in grams/mol)
- GMV is the molar volume (22.4 L/mol at STP)
In this case, the given mass of ozone (O₃) is 0.24 grams. The molar mass of ozone can be calculated by summing the molar masses of each element (oxygen) in the chemical formula.
Molar mass of oxygen (O): 16 g/mol
Since ozone (O₃) has 3 oxygen atoms:
Molar mass of ozone (O₃) = (3 * 16) g/mol = 48 g/mol
Now, substitute the values into the formula:
Volume = (0.24 g * 48 g/mol) / (22.4 L/mol)
Calculate the numerator first:
Numerator = 0.24 g * 48 g/mol = 11.52 g⋅mol
Now divide the numerator by the GMV value:
Volume = 11.52 g⋅mol / 22.4 L/mol
Volume ≈ 0.514 L (rounded to three decimal places)
Therefore, the volume occupied at STP by 0.24 g of ozone is approximately 0.514 liters.