What alkane effuses at 1/5 the rate of He
rate He = 5L/sec
rate x = 1 L/sec
5/1 = sqrt(Mx/4)
Mx = ?
Then CH3(CH2)yCH3
CH3 = 15 and 15*2 = 30 for the two ends.
That leaves Mx-30 for the (CH2)y, solve for y to obtain the formula of the alkane.
To find the alkane that effuses at 1/5 the rate of helium (He), we need to apply Graham's Law of Effusion. Graham's Law states that the rate of effusion of a gas is inversely proportional to the square root of its molar mass.
The molar mass of helium (He) is approximately 4 g/mol. So, if we let x represent the molar mass of the alkane, we can set up the equation:
(1/√x) / (1/√4) = 1/5
To solve for x, we can simplify the equation:
(1/√x) / (1/2) = 1/5
(1/√x) * 2 = 1/5
2/√x = 1/5
Cross-multiplying, we get:
2 = √x/5
10 = √x
x = 100
Therefore, the alkane that effuses at 1/5 the rate of helium is the alkane with a molar mass of approximately 100 g/mol.