Assuming that the mass of sodium methoxide required to prepare 100 grams of 27% methanol solution of sodium methoxide = 100*27% = 27 grams, the mass of methanol required = 100-27 = 73 grams Dissolve 27 grams of sodium methoxide in 73 grams of methanol, you can get 100 g of 27% sodium methoxide in methanol.
How to prepare a methanol solution of sodium methoxide
Jul 18, 2022Leave a message
Send Inquiry