A large cylinder of helium had a small hole through which the helium escaped ata rate of 0.0064 mol/hr. How long would it take for 0.010 mol of CO to leak through a similar hole if it were at the same pressure?

The effusion rate will be inversely proportional to the square root of molecular mass

rate= .0064mol/hr( sqrt MolemassHe/molmassCO)

after you compute that, then time=.010/rate
http://www.chem.tamu.edu/class/majors/tutorialnotefiles/graham.htm

To find the time it would take for 0.010 mol of CO to leak through the hole, we need to use the concept of molar flow rate.

The molar flow rate is the amount of substance passing through a given area per unit time. It is calculated by dividing the amount of substance (in moles) by the time (in hours).

Given that the helium leaked at a rate of 0.0064 mol/hr, we can assume that the molar flow rate is constant for gases at the same pressure and temperature.

Now, we can set up a proportion to find the time it would take for 0.010 mol of CO to leak:

(molar flow rate of helium) : (molar flow rate of CO) = (amount of helium leaked) : (amount of CO to be leaked)

Let's substitute the given values into the equation:

0.0064 mol/hr : (unknown molar flow rate of CO) = 0.010 mol : 1 (since we want to find the time for 0.010 mol of CO)

To find the unknown molar flow rate of CO, we can cross-multiply and solve for it:

0.0064 mol/hr * 1 = (unknown molar flow rate of CO) * 0.010 mol

0.0064 mol/hr = (unknown molar flow rate of CO) * 0.010 mol

Dividing both sides by 0.010 mol:

0.0064 mol/hr / 0.010 mol = (unknown molar flow rate of CO)

0.64 hr⁻¹ = (unknown molar flow rate of CO)

Therefore, the molar flow rate of CO is 0.64 hr⁻¹. This means that 0.010 mol of CO will leak through the hole in 1 hour.

Hence, it would take 1 hour for 0.010 mol of CO to leak through the same hole at the same pressure.