What volume do you want to make. To make 1 liter, you take the 185 g (the molar mass) and dissolve in enough solvent to make the final volume 1 liter.
To prepare a 1 molar solution, you would need to dissolve 180 grams of glucose powder in enough water to make a final volume of 0.5 liters.
To prepare a 1 mole solution of dimethoxyhydroxyacetophenone, you would dissolve 166.21 grams of the compound in enough solvent to make a total volume of 1 liter. Calculate the required weight based on the molar mass of dimethoxyhydroxyacetophenone (C10H12O4).
They're actually exactly the same in that neither of them exists.
To prepare a 1 molar solution of a liquid, you need to dissolve one mole of the solute in enough solvent to make a final volume of 1 liter. Start by calculating the mass of the solute needed using its molar mass. Then dissolve this mass of solute in a volumetric flask with some solvent. Finally, add more solvent to reach the 1 liter mark on the flask and mix well to ensure homogeneity.
To calculate the molar mass of TiCl4, first find the molar mass of each element: Ti (Titanium) has a molar mass of 47.87 g/mol, and Cl (Chlorine) has a molar mass of 35.45 g/mol. Next, multiply the molar mass of each element by the number of atoms in the compound: Ti(Cl)4 = 47.87 + 4(35.45) = 47.87 + 141.8 = 189.67 g/mol. Therefore, the molar mass of TiCl4 is 189.67 g/mol.
What volume do you want to make. To make 1 liter, you take the 185 g (the molar mass) and dissolve in enough solvent to make the final volume 1 liter.
To prepare a 2 M solution of KCl in 1 liter of water, you would need to dissolve 149.5 grams of KCl. This is because the molar mass of KCl is approximately 74.5 g/mol, and 2 moles of KCl are needed to prepare a 2 M solution in 1 liter of water.
To prepare a 1 molar solution, you would need to dissolve 180 grams of glucose powder in enough water to make a final volume of 0.5 liters.
Dilute 1 mL of 0.5 M silver nitrate solution to a total volume of 1 L with water to make a 1 mM silver nitrate solution.
To make a 1 molar solution of sodium azide, you would need to dissolve 65.01 g of sodium azide in water to make 1 liter of solution. Since you have 98 mg of sodium azide, you would need to add enough water to make a final volume of 1 liter to create the 1 molar solution.
Find out the molecular weight of LactoseAdd that many grams of Lactose into a 1000ml volumetric flaskMake up the volume to 1000ml with waterYour 1 Molar solution of Lactose is ready---------------The molar mass of lactose is 342,3 g/mol; the solubility of lactose is 216 g/L at20 0C. Consequently you cannot prepare a molar solution of lactose.
To prepare 0.2M solution of anhydrous sodium thiosulfate (Na2S2O3), you dissolve 24.6g of anhydrous Na2S2O3 in distilled water and dilute it to 1 liter. This is the molar mass method, where molar mass of Na2S2O3 is 158.10 g/mol.
1 molar solution of sugar water contains 342,3 g sucrose.
To make a 0.1 molar solution from a 1.0 molar solution, you would dilute the original solution by a factor of 10. For example, you could mix 1 part of the 1.0 molar solution with 9 parts of solvent (like water) to achieve a final concentration of 0.1 molar.
You have to dissolve 1.00 mol, that is 98.15 g CH3COOK (its molar mass being 98.15 g/mol), in upto 1.000 L.(Suggested procedure: dissolve 98.15 g CH3COOK in not more then 900 mL, homogenize and fill up to exactly 1.000 L by carefully adding the last millilitres water).
Micromolar solution: Suppose 300 is mol wt of compound then 300g in 1000 ml -- it becomes 1M 300,000 mg in 1000 ml ---it is also 1 Molar 1 mg=1000 microgram hence 300,000,000 microgram in 1000 ml ----it is 1 molar now 300,000 microgram in 1 ml ----it is 1 Molar 1 molar=1000 milimolar hence 300,000 microgram in 1 ml -----1000 milimolar 300,000 ----------------------------- 1000,000 micromolar 0.3 microgram --in 1 ml it is 1 micromolar simillarly convert the ml as you want