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To prepare a 0.4 M (mol/L) solution, first, determine the volume of solution you need. Use the formula: moles = molarity × volume. For example, to prepare 1 liter of a 0.4 M solution, you would need 0.4 moles of solute. Weigh the appropriate amount of the solute (using its molar mass), dissolve it in a small volume of solvent, and then dilute to the desired final volume (1 liter) with the solvent.
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What mass of calcium chloride is needed to prepare 2.000 L of a 2.25 M solution?
To calculate the mass of calcium chloride (CaCl₂) needed to prepare a 2.000 L solution at a concentration of 2.25 M, first determine the number of moles required: ( \text{moles} = \text{Molarity} \times \text{Volume} = 2.25 , \text{mol/L} \times 2.000 , \text{L} = 4.50 , \text{mol} ). The molar mass of calcium chloride is approximately 110.98 g/mol, so the mass needed is ( 4.50 , \text{mol} \times 110.98 , \text{g/mol} \approx 497.41 , \text{g} ). Therefore, you would need approximately 497.41 grams of calcium chloride to prepare the solution.
What is the molar concentration of a solution made by dissolving 4.0g of NaOH in enough water to make 1.0 L of solution?
C1V1 = C2V24 x .04 = 1 x V2V2 = (4 x .04)/1= 160mLTherefore the volume of water that needs to be added is 120mL (minus the original volume).
What mass of calcium chloride is needed to prepare 2.85 L of a 1.56 M solution?
To calculate the mass of calcium chloride (CaCl₂) required for a 1.56 M solution, first use the formula: [ \text{mass (g)} = \text{molarity (mol/L)} \times \text{volume (L)} \times \text{molar mass (g/mol)} ] The molar mass of CaCl₂ is approximately 110.98 g/mol. Therefore, for 2.85 L of a 1.56 M solution: [ \text{mass} = 1.56 , \text{mol/L} \times 2.85 , \text{L} \times 110.98 , \text{g/mol} \approx 49.3 , \text{g} ] Thus, about 49.3 grams of calcium chloride is needed.
How do you prepare 1 m M solution of salicylic acid?
To prepare a 1 mM solution of salicylic acid, first calculate the mass required using the formula: mass (g) = molarity (mol/L) × volume (L) × molar mass (g/mol). The molar mass of salicylic acid is approximately 138.12 g/mol. For 1 liter of a 1 mM solution, you would need 0.1381 g of salicylic acid. Dissolve this mass in distilled water and then dilute to a total volume of 1 liter.
How many grams of solute are needed to prepare 2.5L of a 3.0M of AlNO33 solution?
To find the grams of solute needed, use the formula: [ \text{grams} = \text{molarity (M)} \times \text{volume (L)} \times \text{molar mass (g/mol)}. ] The molar mass of Al(NO₃)₃ is approximately 213.00 g/mol. For a 3.0 M solution in 2.5 L: [ \text{grams} = 3.0 , \text{mol/L} \times 2.5 , \text{L} \times 213.00 , \text{g/mol} = 1597.5 , \text{g}. ] Thus, you would need approximately 1597.5 grams of Al(NO₃)₃.
What volume of 18.0 mol/L sulfuric acid is required to prepare 28.2 L of 0.126 mol/L H2SO4?
202.44
Calculate the volume of a 0.05000 mol/l stock solution of CuSO4 (aq) required to prepare 100.0 ml of a 0.005000 mol/l solution?
For starters, you know that 0.05000-mol L − 1 solution of copper(II) sulfate contains 0.05000 moles of copper(II) sulfate, the solute, for every 1 L = 10 3 mL of the solution.
How many grams of iodine would you weigh out to prepare 50 mL of a 0.200 N iodine solution?
To calculate the mass of iodine needed to prepare a 0.200 N solution in 50 mL, you can use the formula: Mass (g) = molarity (mol/L) * volume (L) * molar mass (g/mol). First, convert 50 mL to L (50 mL = 0.050 L). Then, substitute the values: Mass (g) = 0.200 mol/L * 0.050 L * 253.8 g/mol = 2.538 g of iodine. Therefore, you would weigh out 2.538 grams of iodine.
How many grams of K2SO4 would be needed to prepare 4.00 L of a 0.0510 M solution?
To calculate the grams of K2SO4 needed to prepare the solution, use the formula: (molarity) x (volume in liters) x (molar mass of K2SO4). First, calculate the moles of K2SO4 needed: 0.0510 mol/L x 4.00 L = 0.204 mol. Then, find the molar mass of K2SO4: 2*(39.10 g/mol) + 1*(32.07 g/mol) + 4*(16.00 g/mol) = 174.26 g/mol. Finally, multiply the moles by the molar mass: 0.204 mol x 174.26 g/mol ≈ 35.5 grams of K2SO4 are needed.
What is the molar concentration of a solution made by dissolving 4.0g of NaOH in enough water to make 1.0 L of solution?
C1V1 = C2V24 x .04 = 1 x V2V2 = (4 x .04)/1= 160mLTherefore the volume of water that needs to be added is 120mL (minus the original volume).
When was Wouter Mol born?
Wouter Mol was born on 1982-04-17.
What is the concentration of the solution in mol/L?
The concentration of the solution is measured in moles per liter (mol/L).
What mass of calcium chloride is needed to prepare 2.85 L of a 1.56 M solution?
To calculate the mass of calcium chloride (CaCl₂) required for a 1.56 M solution, first use the formula: [ \text{mass (g)} = \text{molarity (mol/L)} \times \text{volume (L)} \times \text{molar mass (g/mol)} ] The molar mass of CaCl₂ is approximately 110.98 g/mol. Therefore, for 2.85 L of a 1.56 M solution: [ \text{mass} = 1.56 , \text{mol/L} \times 2.85 , \text{L} \times 110.98 , \text{g/mol} \approx 49.3 , \text{g} ] Thus, about 49.3 grams of calcium chloride is needed.
How many moles of KCL are contained in 0.538 L of a 2.3M KCL solution?
Since you have you're Molarity and your Liters you can use the formula M= mol/ L 2.3 M = mol / .538 L Multiply both sides by .538 to get the mol alone. mol of KCL = 1.2374
How many grams of NaC2H3O2 are needed to prepare 350 ml of a 2.75 M solution?
First find the number of moles of the sodium bicarbonate by dividing the mass you have(12.5g) by the molar mass of the compound(84.0g) to get .149 moles. Then divide the number of moles (.149mol) by the number of liters of solute (.350L) to get the molarity which is .426 M or .426 mol/L to the correct number of significant figures Molarity = Grams/(Molecular Weight X Volume) Given: grams = 12.5g ; molecular weight of NAHCO3 = 84.00661 g/mol ; volume =350ml or 0.35L Solution Molarity = 12.5 g / ( 84.00661 g/mol x 0.35 L) = 12.5 g / ( 29.4023135 g/mol L ) = 0.425136614 mol L or approx 0.43 mol L
What is the volume in liters of 0.75 mol of methane gas (CH4)?
The molar volume of an ideal gas at standard temperature and pressure (STP) is 22.4 L/mol. Therefore, 0.75 mol of methane gas would occupy 16.8 liters (0.75 mol x 22.4 L/mol = 16.8 L).
How many grams of sodium choride are required to prepare 500.0mL of a 0.100 M solution?
500 mL should contain0.500 (L) * 0.100 (mol / L) * 58.5 (g / mol) = 2.925 g (=grams)
