# How to calculate the van der Waals radius of ethane?

Calculate the van der Waals radius for ethane if $b = 0.0493\ \mathrm{dm^3\ mol^{-1}}$.

$$b = 4 \times {4 \over 3}\pi \times r^3 \times N_\mathrm A$$ $$\implies r = \sqrt[3]{3b\over 16\pi N_\mathrm A}= \sqrt[3]{3 0.049 \times 10^{-3}\over 16\pi N_\mathrm A}\ \mathrm m = 1.70 \times 10^{-10}\ \mathrm m$$

The answer given in the book uses $\displaystyle r =\frac12\sqrt[3]{3b\over 4\pi N_\mathrm A} = 1.35 \times 10^{-10}\ \mathrm m$.

Is my solution correct or is the book's solution correct ?

Edit as per martin's suggestion,

Here are the places I read $b = 4 \times V \times N_a$,

1) :- From the book,

2) :- From the Wikipedia,

(The picture is too large, might have to be opened in a separate tab to be readable)

3) :- The solution in the book,

4) :- Wildcard entry

I think I am missing something very crucial but I don't know what it is.

• Could you explain where your formula for $b$ comes from? Did you take is from a book (which?) or derive it yourself (where did you start?)? Adding a bit more context to the question would probably help us better where the underlying problem is. For example, I can only assume that you make the (not unreasonable) assumption that ethane is a sphere and the volume of a sphere is $V=\frac43\pi r^3$. Where does the 4 come from? – Martin - マーチン Jan 10 '17 at 4:11
• @Martin-マーチン I got the formula from Peter Atkin's physical chemistry. same can be found on Wikipedia . Is there no agreement on what is the value of $b$ ? At different sources I find different values for $b$, here is another value for $b$ i.e $b = V_{molecule} \times N_a$. hyperphysics.phy-astr.gsu.edu/hbase/Kinetic/waal.html . – A---B Jan 10 '17 at 8:53
• Please edit additional information into the post, is the exercise from Atkins book, too? If so, please include a citation. – Martin - マーチン Jan 10 '17 at 9:29
• @Martin-マーチン Ok Martin, I will do that after scanning the book. What do you think about my question in previous comment – A---B Jan 10 '17 at 9:37

• They started out with $$b = N_a \times {4\over 3}\pi \times (2r)^3$$ – A---B Jan 8 '17 at 12:02
• Why ? Isn't $b = \color{red}{4} \times {4 \over 3} \pi r^3 \times N_a$ not $b = \color{red}{8} \times {4 \over 3} \pi r^3 \times N_a$ ???? – A---B Jan 9 '17 at 9:21