If you wanted to add 9.24 x 10-3 mol of 3- bromopentane (M.W. 151.05) to a round
ID: 562928 • Letter: I
Question
If you wanted to add 9.24 x 10-3 mol of 3- bromopentane (M.W. 151.05) to a round bottom flask, how many grams of 3-bromopentane would Enter your answer using two decimal places (12.50), include zeroes, as needed. Include the correct abbreviation for the appropriate unit Answer: If you had added 1.5 mL of methanol (M.W. 32.0, d 0.791 g/mL) to a 25 mL round-bottom flask, how many millimoles of methanol would you have used? Enter your answer using no decimal places (45). Include the correct abbreviation for the appropriate unit Answer:Explanation / Answer
Ans. # Mass of 3-bromopenatne = Moles x Molar mass
= 9.24 x 10-3 mol x (151.05 g/ mol)
= 1.395702 g
= 1.40 g
# Mass of methanol added = Volume x density
= 1.5 mL x (0.791 g/ mL)
= 1.1865 g
Moles of methanol added = Mass / Molar mass
= 1.1865 g/ (32.0 g/ mol)
= 0.037078125 mol ; [1 mol = 1000 millimol]
= 37.08 millimole
Therefore, millimoles of methanol added = 37 millimole