It's simple to dilute a solution, just add more water. If you have one liter of 6 molar solution, and add another 5 liters of water, it becomes a 1 molar solution.
You can also use M1V1 = M2V2 formula for dilution.
For example, if you want to prepare 1 M of solution in 1 litre of water, then how much volume you need from the initial 6 M solution?
Simply use the equation, (6 M) x (V1) = (1 M) x (1 litre) ---> V1 = (1 M) x (1 litre)/(6 M) = 0.167 L or 167 ml
To prepare 1 L of 15 M NaOH solution, you need to dilute the 6 M stock solution. You would need to take 500 mL of the 6 M NaOH stock solution and add water to bring the total volume up to 1 L. This will give you a final concentration of 15 M.
To prepare 600 mL of 1 M HCl solution from a 6 M stock solution, you can use the formula: M1V1 = M2V2, where M1 = concentration of stock solution, V1 = volume of stock solution needed, M2 = final concentration, and V2 = final volume. Plugging in the values: 6 M x V1 = 1 M x 600 mL. Therefore, V1 = 100 mL. Therefore, you would need to measure 100 mL of the 6 M stock solution and dilute it to 600 mL with water to make a 1 M HCl solution.
To prepare a 1 M solution of perchloric acid (HClO4), you would need to dissolve 1 mole of HClO4 in enough water to make a final volume of 1 liter. This can be calculated by using the formula: moles = Molarity (M) x Volume (L). Please handle perchloric acid with caution and use appropriate safety measures while preparing the solution.
6 m HCl refers to the concentration of an Hydrochloric Acid solution. It is 6 molal.A 1 molal solution contains one mole of the solute in 1 kg of solvent. In the abovecase the solvent is water.
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 1 L of 15 M NaOH solution, you need to dilute the 6 M stock solution. You would need to take 500 mL of the 6 M NaOH stock solution and add water to bring the total volume up to 1 L. This will give you a final concentration of 15 M.
To prepare 1 M CaI aqueous solution, dissolve 29.4 g in a total volume of 100 mls, or 294 g in a total volume of 1 liter.
To prepare 600 mL of 1 M HCl solution from a 6 M stock solution, you can use the formula: M1V1 = M2V2, where M1 = concentration of stock solution, V1 = volume of stock solution needed, M2 = final concentration, and V2 = final volume. Plugging in the values: 6 M x V1 = 1 M x 600 mL. Therefore, V1 = 100 mL. Therefore, you would need to measure 100 mL of the 6 M stock solution and dilute it to 600 mL with water to make a 1 M HCl solution.
To prepare a 10 ml of a 25 M HCl solution, you would need to dilute the 1 M HCl solution by a factor of 25. You will need 0.25 ml of the 1 M HCl solution and 9.75 ml of solvent (usually water) to make a total volume of 10 ml for the 25 M HCl solution.
To prepare a 1 M solution of perchloric acid (HClO4), you would need to dissolve 1 mole of HClO4 in enough water to make a final volume of 1 liter. This can be calculated by using the formula: moles = Molarity (M) x Volume (L). Please handle perchloric acid with caution and use appropriate safety measures while preparing the solution.
To prepare a 0.05 M disodium EDTA solution, you would need to dissolve 3.72 grams of disodium EDTA dihydrate (Na2C10H14N2Na2·2H2O) in enough water to make 1 liter of solution.
6 m HCl refers to the concentration of an Hydrochloric Acid solution. It is 6 molal.A 1 molal solution contains one mole of the solute in 1 kg of solvent. In the abovecase the solvent is water.
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.
you need to make 1000 times dilutions, this could be done in multi-steps: transfer 1 ml 0.1 M HCl into 100 ml volumetric flask and complete volume with water --------(1) from solution (1) transfer 2 ml into 20 ml volumetric falsk and complete volume with water, this is 0.0001 M HCL.
To prepare a 0.50 M acetic acid solution, you would need to dilute the 2.5 M stock solution. By using the formula M1V1 = M2V2, you can calculate the volume of the stock solution needed as: (0.5 M)(100.0 mL) = (2.5 M)(V2), where V2 is the volume of the stock solution needed. So, V2 = (0.5 M x 100.0 mL) / 2.5 M = 20.0 mL. Therefore, you would need 20.0 mL of the 2.5 M stock solution to prepare the desired 100.0 mL of 0.50 M acetic acid solution.
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.
A solution of HCl is highly dissociated into ions, A 0.000001 M solution (1 x 10-6) has a pH of 6 ... close to neutral. A 0.001 M solution (1 x 10-3) has a pH of 3 ... more concenterated, but still not a really concentrated solution. A 0.1 M solution (1 x 10-1) has a pH of 1 ... even more concentrated. showing it is more acidic.