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What mass in grams of solute is needed to prepare 70.0 mL of 2.70 M C6H12O6?
Wiki User
∙ 8y ago
Molar mass of C6H12O6 is 180 g/ mole. 70 ml is 0.070 liters
2.70 M is 2.7 moles/liter
So... 2.70 mol/L x 0.070 L x 180g/mole = 34.02 g = 34.0 g (3 sig figs)
Wiki User
∙ 8y ago
Wiki User
∙ 8y agoThe molar mass of glucose is 180,16 g. 2,7 M is 486,43 g.
The amount needed for 70 mL is 34,05 g.
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If 154 grams of glucose are dissolved to yield one liter of solution what is the molarity of the 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 ---------------------------
What is the molarity of an aqueous solution containing 32.8 g of C6H12O6 in 1.0 L of solution?
Find moles of glucose. 32.8 grams C6H12O6 (1 mole C6H12O6/180.156 grams) = 0.18206 moles C6H12O6 Molarity = moles of solute/Liters of solution M = 0.18206 moles C6H12O6/1.0 L = 0.18 M C6H12O6
What mass of oxygen in grams is in 147.2 grams of glucose?
147.2 grams C6H12O6 (1 mole C6H12O6/180.156 grams)(6 moles O/1 mole C6H12O6)(16.0 grams/1 mole O) = 78.4 grams oxygen ----------------------------
If you are giving a quantity of solute in grams and need to convert it to moles divide the grams by?
The molecular weight. So you'll have to calculate that for the solute first. The molar mass of the solute, which is measured in grams/mole.
How many grams are equal to 450 milligrams?
Molarity = moles of solute/Liters of solution ( 450 ml = 0.450 liters) 5M C6H12O6 = moles C6H12O6/0.450 liters = 2.25 moles C6H12O6 (180.156 grams/1 mole C6H12O6) = 405.351 grams of glucose ( you do significant figures )
If 154 grams of glucose are dissolved to yield one liter of solution what is the molarity of the 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 ---------------------------
What is the molarity of an aqueous solution containing 32.8 g of C6H12O6 in 1.0 L of solution?
Find moles of glucose. 32.8 grams C6H12O6 (1 mole C6H12O6/180.156 grams) = 0.18206 moles C6H12O6 Molarity = moles of solute/Liters of solution M = 0.18206 moles C6H12O6/1.0 L = 0.18 M C6H12O6
How many grams of C6H12O6 are needed to form 7.50g of C2H5OH?
14.7 g
How many grams needed from water to prepare 160 grams from potassium acetate 5 percent w?
160 g (solution) - [5/100*160] g (solute) = 152 g (solvent) water
How many grams of glucose are needed to make one liter of a 1.0 molar solution?
186g C6H12O6 Use a concentration formula. Molarity - moles of solute/liters of solution. molarity = 1 moles of solute= x liters= 1 solve the equation and x= 1. One mole of glucose is equal to 186 grams.
What mass of oxygen in grams is in 147.2 grams of glucose?
147.2 grams C6H12O6 (1 mole C6H12O6/180.156 grams)(6 moles O/1 mole C6H12O6)(16.0 grams/1 mole O) = 78.4 grams oxygen ----------------------------
How many grams of KCl are required to prepare 2.25L of a 0.75m KCl solution?
Molarity = moles of solute/Liters of solution 0.75 M KCl = moles KCl/2.25 Liters = 1.6875 moles KCl (74.55 grams/1 mole KCl) = 126 grams of KCl needed
How many grams of carbon dioxide would be produced if 45 g of C6H12O6 competely reacted with oxygen?
C6H12O6 + 6O2 --> 6CO2 + 6H2O 45 grams C6H12O6 (1 mole C6H12O6/180.156 grams)(6 moles CO2/1 mole C6H12O6)(44.01 grams/1 mole CO2) = 66 grams carbon dioxide produced ==========================
How many moles are contained in 15.8 grams of C6H12O6?
15.8 grams of C6H12O6 is equivalent to 0.087 moles.
How many grams of solute are needed to prepare 250.0ml of 0.125 M CuSO4?
M means mol/litre, you have 250ml. so divide .125 by 1000, x250 to get the moles. then multiply by molecular weight of copper sulphate
How many grams of CaCl2 are needed to prepare 2.00 l if 0.700 M CaCl2 solution?
The gram formula mass of CaCl2 is 110.99. By definition, each liter of 0.700 M CaCl2 contains 0.700 gram formula masses of the solute. Therefore, 2.00 liters of such solution contain 1.400 formula masses of the solute, or 155 grams, to the justified number of significant digits.
How may grams of KNO3 should be used to prepare a 2.0L of a 0.50 M solution?
Get moles by; Molarity = moles of solute/Liters of solution 0.50 M KNO3 = moles KNO3/2.0 L = 1.0 mole KNO3 Now find grams of 1.0 mole KNO3 1.0 mole KNO3 (101.11 grams/1 mole KNO3) = 101.11 grams KNO3 needed call it 100 grams
