Answer:
58 g/mol
Explanation:
According to Graham's law, the rate of diffusion of a gas (r) is inversely proportional to the square root of its molar mass (M). Butane's rate of diffusion is 3.8 times slower than that of helium, that is, rButane = rHe/3.8, or rHe/rButane = 3.8. Then,
[tex]\frac{rHe}{rButane} =3.8=\sqrt{\frac{M(Butane)}{M(He)} } =\sqrt{\frac{M(Butane)}{4.00g/mol} } \\M(Butane)=3.8^{2} \times 4.00g/mol\\M(Butane)=58g/mol[/tex]
Answer:
The molar mass of butane ≈ 58 g/mol
Explanation:
Step 1: Data given
diffusion rate of butane is 3.8 slower than helium
Let's say the diffusion rate of butane is 1 and helium its diffusion rate = 3.8
Step 2: Calculate the molar mass of butane
Rate 1 / rate 2 = √(MM2 / MM1)
⇒ with rate 1 = the rate of diffusion of helium = 3.8
⇒ with rate 2= the rate of diffusion of butane = 1
⇒ with MM1 = the molar mass of helium = 4.00 g/mol
⇒ with MM2 = the molar mass of butane = TO BE DETERMINED
rate helium / Rate butane = √(MM of butane / MM helium)
3.8 /1 = √(MMbutane/4)
3.8² *4.00 = MMbutane
MM of butane = 57.8 g/mol ≈58 g/mol
The molar mass of butane ≈ 58 g/mol
