Chemistry

# How many grams of CH3OH is needed to make 0.211m solution in 479g of water?

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## Related Questions

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

CH3OH + NH3 &rarr; CH3NH2 + H2O 32 g of methanol reacts with 17.0 g ammonia 21 g of CH3OH react with 11.0 g NH3

9.32*1024 (molec's CH3OH) / 6.022*1023 (molec's.mol&minus;1 CH3OH) * 32.04 (g mol&minus;1 CH3OH) = 495.8 g = 496 g CH3OH

9,85x10 to the 24 of methanol (CH3OH) molecules have a mass of 52,405 g.

9.32*1024 (molec's CH3OH) / 6.022*1023 (molec's.mol&minus;1 CH3OH) * 32.04 (g mol&minus;1 CH3OH) = 495.8 g = 496 g CH3OH

7.5 grams are needed because each milliliter is 1.50 grams, to get 5.00 mls, the calculation would be 5x1.50=7.5.

HCl has a molar mass of 36.461 grams per mole. This means that 72.922 grams of HCl are needed per liter of water to make a solution that has a concentration of 2M.

how many grams of calcium nitrate are needed to make a 500ml volume of a .5 molar solution

How many grams of KHP are needed to exactly neutralize 36.7 mL of a 0.328 M barium hydroxidesolution

9.18x10^24 molecules CH3OH x 1 mole/6.02x10^23 molecules x 32 g/mole = 488 g (to 3 sig figs)

If your solution is a total of 414g and 3.06% of it needs to be NaCl, then you just take 414 x .0306 = grams of NaCl. The rest of the grams will be from other species in the solution.

Molarity = moles of solute/Liters of solution 3 M KBr = moles KBr/1 liter = 3 moles KBr (119 grams/1 mole KBr) = 357 grams needed

Molarity = moles of solute/volume of solution 1.50 M = X Moles/125ml = 187.5 millimoles, which is 0.1875 moles The molar mass of CaCl2 = 110.98 grams 0.1875 moles CaCl2 (110.98g/1mol) = 20.8 grams needed

Weight to volume solution is calculated by the weight of the solute in grams divided by the volume of the solution in milliliters. Assuming that the solvent is water, then then 100 grams of glucose is needed.

Put 100 grams in a beaker and and around 500 mls of water until it dissolves, then top up the beaker to a liter. That is your 10% solution. The percentage solution is a ratio of the weight of the compound to the weight of the final solution.

400 mls would require 40g of glucose for a 10% solution and thus 20g for a 5% solution.

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