Chapter 13 - The Chemistry of Solutes and Solutions - Problem Solving Practice 13.6 - Page 582: a

$1.93\,g$

Volume= $250\,mL= 0.250\,L$ Molarity=$0.0750\,M$ Number of moles $n=Molarity\times Volume$ $=0.0750\,M\times0.250\,L=0.01875\,mol$ Molar mass of NaBr= 102.894 g/mol Mass in grams= number of moles$\times$molar mass $=0.01875\,mol\times102.894\,g/mol=1.93\,g$