A certain gas diffuses four times as quickly as oxygen. The molecular weight of the gas is
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According to Graham's Law, the rates of diffusion are related thus:
rate1 / rate2 = sqrt(M2 / M1)
So let's make oxygen "Gas 2" and we'll call the mystery gas "Gas 1". Since Gas 1 diffuses 4 times as fast as oxygen, we know that:
rate1 / rate2 = 4
Substituting that into the Graham's Law equation:
4 = sqrt(M2 / M1)
The molar mass (M2) of oxygen gas, O2, is 32 g/mol.
4 = sqrt(32 / M1)
Square both sides.
16 = (32 / M1)
M1 = 32 / 16 = 2
The only gas that has a molar mass of 2 g/mol is hydrogen gas, H
rate1 / rate2 = sqrt(M2 / M1)
So let's make oxygen "Gas 2" and we'll call the mystery gas "Gas 1". Since Gas 1 diffuses 4 times as fast as oxygen, we know that:
rate1 / rate2 = 4
Substituting that into the Graham's Law equation:
4 = sqrt(M2 / M1)
The molar mass (M2) of oxygen gas, O2, is 32 g/mol.
4 = sqrt(32 / M1)
Square both sides.
16 = (32 / M1)
M1 = 32 / 16 = 2
The only gas that has a molar mass of 2 g/mol is hydrogen gas, H
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