Answer to Question #324632 in Chemistry for Ace

Question

How many grams of NaBr (molar mass = 102.9 g/mol) would be needed to prepare 700 ml of 0.230 M NaBr solution? 

Answer 1

C_M (NaBr) = n / V

n (NaBr) = C_M x V = 0.230 x 0.700 = 0.161 mol

n = m / M

m = n x M

M (NaBr) = 102.9 g/mol

m (NaBr) = 0.161 x 102.9 = 16.6 g