Chem120 Final Exam Verified And Updated Question And Answers Version 2

CHEM120 FINAL EXAM VERIFIED AND UPDATED QUESTION AND ANSWERS VERSION 2 Multiple Choice 1.What is the molarity of a solution containing 55.8 g of MgCl2 dissolved in 1.00 L of solution? M= moles of solute liters of solution. We are given the mass of our solute so we just need to convert it to moles of MgCl2 via the MM. 55.9 g MgCl2 x 1 mol MgCl2 =0.586111 mols MgCl2 95.2119 g MgCl2 Now we can plug this into our equation for molarity M = 0.587111 mols MgCl2 = 0.586 M MgCl2 `1.00 L a) 55.8 M b) 1.71 M c) 0.586 M d) 0.558 M 2.What is the mass in grams of Mg(NO3)2 present in 145 mL of a 0.150 M solution of Mg(NO3)2? We know the molarity and the volume, but we need to find the moles. M= moles of solute liters of solution. (volume) First convert mLL. 145 mL 0.145 L (divide by 1000) so we can solve this equation for moles. Moles= (M) x (Volume) moles Mg(NO3)2 = (0.150 moles) x (0.145 liters) = 0.02175 moles Mg(NO3)2 liter Now we can convert the number of moles to grams via the molar mass of Mg(NO3)2.0.02175 moles Mg(NO3)2