how many grams of glucose must be added to 525g of 2.5 percent leg mass glucose solution?and give the furmela?
A student placed 15.5 g of glucose (C6H12O6) in a volumetric flask, added enough water to dissolve the glucose by swirling, then carefully added additional water until the 100. mL mark on the neck of the flask was reached. The flask was then shaken until the solution was uniform. A 40.0mL sample of this glucose solution was diluted to 0.500 L. How many grams of glucose are in 100. mL of the final solution?
You need 0,9 glucose.
No, not exactly.Mass mass percent concentration measures grams of solute per 100 grams of solution (= solvent + solute)Example:58.5 g NaCl (solute) added to941.5 g H2O (solvent) gives you1000.0 g solution of 5.85% NaCl (= 100%*58.5/1000.0),which is about a 1.0 mol/L NaCl solution.
Molarity = moles of solute/Liters of solution Find moles glucose, which is molecular formula------C6H12O6 154 grams C6H12O6 (1 mole C6H12O6/180.156 grams) = 0.8548 moles C6H12O6 Molarity = 0.8548 moles C6H12O6/1 Liter = 0.855 M glucose ---------------------------
The percent concentration is 13,75 %.
40.8 grams
200 grams/1,000 mL x 100= 20%
400 mls would require 40g of glucose for a 10% solution and thus 20g for a 5% solution.
12*(90%)=10.8=X*.25 10.8/.25=X=43.2 43.2-12=31.2 31.2 grams of water must be added
195 grams.
A student placed 15.5 g of glucose (C6H12O6) in a volumetric flask, added enough water to dissolve the glucose by swirling, then carefully added additional water until the 100. mL mark on the neck of the flask was reached. The flask was then shaken until the solution was uniform. A 40.0mL sample of this glucose solution was diluted to 0.500 L. How many grams of glucose are in 100. mL of the final solution?
No, not exactly.Mass mass percent concentration measures grams of solute per 100 grams of solution (= solvent + solute)Example:58.5 g NaCl (solute) added to941.5 g H2O (solvent) gives you1000.0 g solution of 5.85% NaCl (= 100%*58.5/1000.0),which is about a 1.0 mol/L NaCl solution.
You need 36,03 g glucose.
You need 0,9 glucose.
Molarity = moles of solute/Liters of solution Find moles glucose, which is molecular formula------C6H12O6 154 grams C6H12O6 (1 mole C6H12O6/180.156 grams) = 0.8548 moles C6H12O6 Molarity = 0.8548 moles C6H12O6/1 Liter = 0.855 M glucose ---------------------------
The percent concentration is 13,75 %.
Very simply 1g of ammonium thiocyanate and 100g (100ml) of water!