How many grams of solute are present in 615 ml of 0.630 m kbr?

Molarity is the number of moles of solute in 1 L of solution

molarity of KBr solution is 0.630 M

this means that there there are 0.630 moles of KBr in 1 L solution

then in 615 mL number of KBr moles are - 0.630 mol/L x 0.615 L = 0.387 mol

mass of KBr is - molar mass x number of moles

KBr molar mass is 119 g/mol x 0.387 mol = 46.1 g

mass of KBr is 46.1 g