A 1.20-L container contains 1.10 g of unknown gas at STP.What is the molecular weight of the unknown gas?
1.20 L*1 mole/22.4 L= 0.05357 moles
L units cx out.
Then use your given mass of
1.10 g/0.05357 moles=20.5 g/mole
To find the molecular weight of the unknown gas, we need to use the ideal gas law equation PV = nRT, where P is the pressure, V is the volume, n is the number of moles, R is the ideal gas constant, and T is the temperature.
At STP (Standard Temperature and Pressure), the temperature is 273.15 K and the pressure is 1 atmosphere (atm).
Given:
Volume (V) = 1.20 L
Mass of gas (m) = 1.10 g
First, we need to find the number of moles of the gas using the ideal gas law equation:
PV = nRT
Rearrange the equation to solve for n:
n = PV / RT
Substitute the known values into the equation:
n = (1atm * 1.20L) / (0.0821atm·L/mol·K * 273.15K)
n ≈ 0.05 moles
Next, we can calculate the molecular weight (molar mass) of the gas using the formula:
Molecular weight (MW) = Mass (m) / Number of moles (n)
Substitute the known values into the equation:
MW = 1.10g / 0.05mol
MW = 22 g/mol
Therefore, the molecular weight of the unknown gas is approximately 22 g/mol.
1.10g/mole
hmmmm. At STP a mole of any gas occupies 22.4 liters, so
1.20/?moles=22.4/1mole
solve for ? moles, then
?moles=1.10g/molmass solve for molmass.