How many grams of solute are present in 445mL of 0.880 M KBr

Question from my chemistry class that I have no idea how to do. Please help!
Thanks

0.880 mols / L KBr * .445 L = 0.392 mols KBr

0.392 mols * (119 g/mol) = 46.6 grams of KBr


First you find the mols of KBr by multiplying the concentration by the volume you have. Then multiply the mols of KBr you calculated by the molar mass of KBr to find the grams of solute.

K = 39.09 g/mol
Br = 79.90 g/mol

445mL * 0.880 mol/ 1000mL * 118.99 g/mol = 46.596 g KBr

You can also guestimate. You have ~.5L at ~.8mol/L. So you'll have somewhere under 1/2 a mol worth of substance.