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How many grams of CH3OH must be added to water to prepare 150 mL of a solution that is 1.5 M CH3OH?
Wiki User
∙ 6y ago
7,207 g methanol are needed.
Wiki User
∙ 6y ago
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How many grams of ch3oh must be added to water to prepare 150 ml of a solution that is 2.0 m ch3oh?
The answer is 9.6 grams.
What is the mass of 3.62 X 1024 molecules of CH3OH?
192.678 grams CH3OH
- How many grams of water should be added to prepare a 2% solution by mass with 16.5g of MgCl….A) 402.5B) 808.5C) 155D) 400E) 82.5?
E
How many grams of I2 should be added to 750 g o C Cl4 to prepare a 020 m solution?
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
What is the mass in grams of 9.32x1024 molecules of methanol CH3OH?
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
How many grams of ch3oh must be added to water to prepare 150 ml of a solution that is 2.0 m ch3oh?
The answer is 9.6 grams.
How many grams of CH3OH is needed to make 0.211m solution in 479g of water?
The answer is 3,211 g.
How do you prepare 30 percent sucrose solution?
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.
What is the mass of 3.62 X 1024 molecules of CH3OH?
192.678 grams CH3OH
How many grams of KMnO4 should be used to prepare 2L of a 5M solution?
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.
- How many grams of water should be added to prepare a 2% solution by mass with 16.5g of MgCl….A) 402.5B) 808.5C) 155D) 400E) 82.5?
E
How many grams of I2 should be added to 750 g o C Cl4 to prepare a 020 m solution?
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
What is the mass in grams of 9.32x1024 molecules of methanol CH3OH?
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
How many grams of sulfosalicylic acid are required to prepare 1 liter of 3 solution?
30 grams
What is the mass in grams of 9.03 X 1024 molecules of methanol CH3OH?
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
How many grams of active ingredient will you need to prepare 2.5 gallons of a 45.6 percent solution?
4314.9 grams
What is the molality of a solution of chlorine and water is 0.0362 m how much chlorine in grams was used to prepare this solution?
8.9g
