The answer is 9.6 grams.
take the formula weight for ch3oh...roughly 36 multiply by 2, and then multiply that by .15 since molarity is in liters and not in ml like in the problem.
answer 10.8 grams ch3oh
7,207 g methanol are needed.
192.678 grams CH3OH
E
If this is .020m solution, you need 3.81g If this is .20m solution, you need 38.1g If this is 20m solution, you need 3810g
9.32*1024 (molec's CH3OH) / 6.022*1023 (molec's.mol−1 CH3OH) * 32.04 (g mol−1 CH3OH) = 495.8 g = 496 g CH3OH
7,207 g methanol are needed.
The answer is 3,211 g.
See the two Related Questions to the left for the answer.The first is how to prepare a solution starting with a solid substance (and dissolving it). The second question is how to prepare a solution by diluting another solution.
192.678 grams CH3OH
To prepare 2L of a 5M solution, you should put in 4.6575grams of KMnO4. It is important to make sure that you add them in that order. K should be added first, then Mn.
E
If this is .020m solution, you need 3.81g If this is .20m solution, you need 38.1g If this is 20m solution, you need 3810g
9.32*1024 (molec's CH3OH) / 6.022*1023 (molec's.mol−1 CH3OH) * 32.04 (g mol−1 CH3OH) = 495.8 g = 496 g CH3OH
30 grams
9.32*1024 (molec's CH3OH) / 6.022*1023 (molec's.mol−1 CH3OH) * 32.04 (g mol−1 CH3OH) = 495.8 g = 496 g CH3OH
4314.9 grams
8.9g