How many grams of cabr2 are needed to prepare 4.65 l of a 7.65 m solution?

Molarity is defined as the number of moles of solute dissolved in 1 L of solution 
the molarity of the solution to be prepared is 7.65 M
the volume to be prepared of CaBr₂ is 4.65 L
the number of CaBr₂ moles are - 7.65 mol/L x 4.65 L = 35.6 mol
the molar mass of CaBr₂ - 200 g/mol
mass of 35.6 mol - 35.6 mol x 200 g/mol = 7120 g
mass required of CaBr₂ to make the solution is 7120 g

The answer should be c.