Here are the solutions:
1) Calculate the molarity of 25.0 grams of KBr dissolved in 750.0 mL.
Note the change from mL to L.
2) 80.0 grams of glucose (C6H12O6, mol. wt = 180. g/mol) is dissolved in enough water to make 1.00 L of solution. What is its molarity?
