Given:

Concentration of CaBr\(_2\) = 20 gdm\(^{-3}\)

  CaBr\(_2\) ⇌  Ca\(^{2+}\)  + 2 Br\(^-\)

 1mole 2moles

(a) Molarity (Molar concentration) of CaBr\(_2\) = \(\frac{Mass concentration}{Molar mass}\)

Molar mass of  CaBr\(_2\) = 40 + (80 x 2) = 200g/mol

 Molarity of CaBr\(_2\) = \(\frac{20}{200}\)

 = 0.100M

(b) Since 1mole of  CaBr\(_2\) produces 2moles of Br\(^-\)

 Molarity of 2Br\(^-\) = 2 x 0.100

  = 0.200M

The molarity of the solution with respect to calcium bromide and bromide ions are 0.10 and 0.20 respectively.