What will be the molecular diameter of Helium if van der Waal’s constant, b = 24 ml mole–1 ?

The Van der Waals constant 'b' is related to the size of the gas molecules. It is four times the actual volume occupied by one mole of the gas molecules.

Given, b = 24 ml/mole

We know that, b = 4 * volume of one mole of gas molecules

So, volume of one mole of gas molecules = b/4 = 24 ml/mole / 4 = 6 ml/mole

Now, we know that one mole of any substance contains Avogadro's number (6.022 x 10^23) of entities.

So, volume of one molecule = volume of one mole of molecules / Avogadro's number = 6 ml/mole / 6.022 x 10^23 molecules/mole = 9.96 x 10^-26 ml/molecule

We can convert this volume to cm^3 (since 1 ml = 1 cm^3), so volume of one molecule = 9.96 x 10^-26 cm^3/molecule

Now, we know that the molecules are spherical in shape, so we can use the formula for the volume of a sphere to find the diameter.

Volume of a sphere = 4/3 * π * (d/2)^3

Rearranging for d (diameter), we get d = [(6 * volume) / π]^(1/3)

Substituting the volume of one molecule into this equation, we get:

d = [(6 * 9.96 x 10^-26 cm^3) / π]^(1/3) = 2.6 x 10^-8 cm

So, the molecular diameter of Helium is approximately 2.6 x 10^-8 cm.