An element E of atomic mass 150g/mol crystallises into an FCC lattice of edge length 600pm the no. of atom present in 15g of element is ax10^​22 . what is  the value of a ?

Dear Student,

The number of atoms can be simply found by finding out the number of moles and then multiplying it by Avogadro's number. We have been given the weight of the compound and the atomic mass.
So,
$$\begin{gathered} Number of moles = \frac{Given mass}{molar mass}=\frac{15}{150}=0.1 mol \\ Number of atoms = Number of moles \times N_A=0.1 \times 6.022 \times {10}^{23}=6.022\times {10}^{22} \end{gathered}$$
Hope this information will clear your doubts about this topic.

If you have any doubts just ask here on the ask and answer forum and our experts will try to help you out as soon as possible.

Also CLICK HERE and check out the POWER TIPS for JEE Advanced 2017.

Regards
Thus, the value of a is 6.022.