How many grams of KBr are required to make 350. mL of a 0.115 M KBr solution?

0.115 M means that 0.115 moles of KBr are contained in a volume of 1000 ml, therefore a volume of 350 ml will have (0.115 * 0.35) = 04025 moles

From the formula of molarity moles = molarity * volume in liters

1 mole of KBr is equivalent to 119 g

Therefore, the mass = 0.04025 * 119 g = 4.79 g