The total contribution of corner ions to the unit cell of $\mathrm{CsCl}$ is____________

In CsCl, the unit cell is cubic with Cl at the corners and Cs at the body centre. The corner atoms effectively contribute one atom to the cell 

There are only eight corner lattice points present in a simple cubic crystal structure and each point is shared by eight unit cells. As a result, the proportion of an atom occupying a corner lattice to the unit cell is $\frac{1}{8}$
Therefore the total number of atoms available in a simple cubic unti cell $\mathrm{n}=8\times \frac{1}{8}=1$

Hence, the required answer is 1.