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58.44g/(mol) The formula of Sodium chloride is NaCl. The atomic mass of sodium is 22.99 g/(mol), and the atomic mass of chlorine is 35.45g/(mol).
A 1M solution that contains 1liter is prepared by weighing 58.44 grams of NaCl and placing this amount of salt in a 1 Liter volumetric flask and then filling the flask with distiller water to the graduation mark. This question requires an understanding of solution concentration which is expressed as molarity (M). Molarity = moles of solute/liters of solution. Since you cannot measure moles ...
"58.44277 g/mol" Sodium chloride, also known as table salt, is an ionic compound composed of sodium cations, "Na"^(+), and chloride anions, "Cl"^(-) that has the formula unit "NaCl". The molar mass of an anionic compound tells you what the mass of one mole of formula units of that respective compound is. You know that sodium chloride's formula unit contains one sodium atom one chlorine atom ...
How would you determine the formula weight of NaCl? Here’s the best way to solve it. Solution. 100 % ...
Question: You have a 0.025M stock solution of NaCl (Formula Weight: 58.4g/mole), a 0.5M stock solution of glucose (Formula Weight; 180.2g/mole), and a bottle of solid Tris base (Formula Weight: 121.1g/mole). How would prepare (be specific) 50mL of a single solution containing 150mM Tris, 25mM glucose, and 5mM NaCl?
How many grams of solid NaCl (Formula Weight: 58.44 g/mole) are required to prepare 150 mL of a 250mM solution? If you dissolve 15mg glucose (Formula Weight: 180.16g/mole) in 10mL of water (all answers must be presented in scientific notation format): What is the concentration of glucose in Mm and mu M?
The formula weight of NaCl is 58.4 g/mole. Wite the description as if you were teaching to your classmate how to make this solution in the lab, showing the necessary calculations and including the units and correct measuring devices that they would need to make it properly from start to finish. You are given a 2 M NaCl stock solution.
Question: You have a 0.15M stock solution of NaCl (Formula Weight: 58.4g/mole), a 0.05M stock solution of glucose (Formula Weight; 180.2g/mole), and a bottle of solid Tris base (Formula Weight: 121.1g/mole). How would prepare (be specific) 75mL of a single solution containing 50mM Tris, 5mM glucose, and 2.5mM NaCl.
It is imperfect to define molecular weight for the substance having no molecular entity as in the case of purely ionic compound like NaCl. So Formula weight is more perfect for it. Some compounds do not exist as individual molecules. For example, there is no such thing as a molecule of sodium chloride. That's why it is appropriate to use the term formula weight for this type of compound ...
6. (25 points) You have a 0.025 M stock solution of NaCl (Formula Weight: 58.4 g / mole), a 0.15 M stock solution of glucose (Formula Weight; 180.2 g / mole), and a bottle of solid Tris base (Formula Weight: 121.1 g / mole). How would prepare (be specific) 50 mL of a single solution containing 20 mM Tris, 30 mM glucose, and 2.5 mM NaCl. 7.