158g of a gas at stp occupies a volume of 50.00dm^3 ..what is the relative moat mass of the gas?
(G M V at stp =22.4dm ^3 mold )
22.4 dm^3/one molar mass=50/molar mass
molar mass= 22.4/50=1.1.2 molar masses
so one molar mass= 158/1.12 grams
To find the relative molar mass of the gas, we need to use the ideal gas law equation:
PV = nRT
Where:
P = pressure
V = volume
n = moles of gas
R = ideal gas constant (0.0821 L·atm/mol·K at STP)
T = temperature (in Kelvin)
At STP (standard temperature and pressure), the pressure is 1 atm and the temperature is 273 K.
First, we need to convert the given volume from dm^3 to L (liters), since the ideal gas constant is given in liters:
1 dm^3 = 1 L
So 50.00 dm^3 = 50.00 L
Now we can rearrange the ideal gas law equation to solve for moles (n):
n = PV / RT
Plugging in the values:
P = 1 atm
V = 50.00 L
R = 0.0821 L·atm/mol·K
T = 273 K
n = (1 atm) (50.00 L) / (0.0821 L·atm/mol·K) (273 K)
n ≈ 1.94 mol
Next, we need to calculate the molar mass of the gas. Since we know the mass of the gas is 158 g and the number of moles is 1.94 mol, we can divide the mass by the number of moles:
Molar Mass = Mass / Moles
Molar Mass = 158 g / 1.94 mol ≈ 81.44 g/mol
Therefore, the relative molar mass of the gas is approximately 81.44 g/mol.