According to I.P. 1996
Use the eq'n moles = [conc] X vol(mL) / 1000 Algebraically rearragne [conc] = moles X 1000 / vol(mL) [conc] = 0.002 X 1000 / 2000 mL [conc] = 0.002 X 1/2 [conc] = 0.001 mol L^-1 or 0.001 M
To make 1 liter of 0.05 M HCl solution, you would need to dilute concentrated HCl. The concentration of the concentrated HCl would depend on its specific concentration, but you would typically need around 100 mL of concentrated HCl, assuming it is a standard 12 M concentration, to make the desired dilution.
To prepare 100 mL of 1.0 M HCl from a 3.0 M stock solution, you can use the formula: (M_1V_1 = M_2V_2). Solving for V1: (3.0 M)(V1 mL) = (1.0 M)(100 mL), thus V1 = 33.3 mL. So, you would need to measure out 33.3 mL of the 3.0 M HCl solution and then dilute it to 100 mL to obtain 1.0 M HCl.
To find the volume of 12.0 M HCl needed to make 75.0 mL of 3.50 M HCl, you can use the formula C1V1 = C2V2. Substituting in the given values, you will find that approximately 21.9 mL of 12.0 M HCl is required.
85 mL of HCl may be taken to prepare a normal solution because it contains a standardized amount of solute molecules (HCl) that will react with a known amount of solvent (usually water) to form a solution of known concentration. This allows for accurate and reproducible measurements and calculations in chemical experiments.
dilute 1.7 ml of Conc. HCl to 1000 ml with water
Use the eq'n moles = [conc] X vol(mL) / 1000 Algebraically rearragne [conc] = moles X 1000 / vol(mL) [conc] = 0.002 X 1000 / 2000 mL [conc] = 0.002 X 1/2 [conc] = 0.001 mol L^-1 or 0.001 M
To make 1 liter of 0.05 M HCl solution, you would need to dilute concentrated HCl. The concentration of the concentrated HCl would depend on its specific concentration, but you would typically need around 100 mL of concentrated HCl, assuming it is a standard 12 M concentration, to make the desired dilution.
To prepare 100 mL of 1.0 M HCl from a 3.0 M stock solution, you can use the formula: (M_1V_1 = M_2V_2). Solving for V1: (3.0 M)(V1 mL) = (1.0 M)(100 mL), thus V1 = 33.3 mL. So, you would need to measure out 33.3 mL of the 3.0 M HCl solution and then dilute it to 100 mL to obtain 1.0 M HCl.
To find the volume of 12.0 M HCl needed to make 75.0 mL of 3.50 M HCl, you can use the formula C1V1 = C2V2. Substituting in the given values, you will find that approximately 21.9 mL of 12.0 M HCl is required.
85 mL of HCl may be taken to prepare a normal solution because it contains a standardized amount of solute molecules (HCl) that will react with a known amount of solvent (usually water) to form a solution of known concentration. This allows for accurate and reproducible measurements and calculations in chemical experiments.
Ah, preparing a 0.02 M solution of HCl is a wonderful journey. Simply measure out the correct amount of hydrochloric acid and dilute it with water until you reach the desired concentration. Remember to handle chemicals with care and always wear appropriate safety gear. Happy experimenting, my friend!
100 M HCl don't exist.
Use v1*C1=v2*C2,so:25.00 mL * [HCl] = (17.65 mL * 0.110 M)thus:[HCl] = (17.65 mL * 0.110 M) / 25.00 mL = 0.07766 M = 0.0777 mol HCl / L (rounded to 3 significants)
To find the concentration of HCl, you can use the formula: moles of NaOH = moles of HCl. From the given information, you can calculate the moles of NaOH used to neutralize the acid. Then, use the volume and concentration of NaOH to determine the concentration of HCl.
To make 0.25N HCl from 1.00N HCl, you would need to dilute the 1.00N HCl solution by adding three parts of water for every part of the original solution. For example, you can mix 1 mL of 1.00N HCl with 3 mL of water to obtain 0.25N HCl solution.
For 1 Liter you should take 88.8 mL concentrated HCL (be carefull) and add it to 900 mL water, and finally - after mixing- fill this up to 1000 mL with distilled water. (REMEMBER: ALWAYS ADD ACID TO WATER! -1N in 250 ml : add 22.2 mL to about (or less than) 225 mL water and dilute to final volume = 0.25L