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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!
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RossTo prepare a 0.02 M solution of HCl, you would need to dilute concentrated HCl solution with a calculated amount of water. For example, to prepare 1 liter of 0.02 M HCl solution, you would mix 4.2 ml of concentrated HCl (12 M) with 995.8 ml of water. Always add acid to water slowly and with stirring to avoid splattering.
0.02M=20mM(multiply x1000)
From a 1M stock HCL I shall prepare 100ml of 20mM HCl by the following equation
V1N1=V2N2
100mlx20mM=1000mMx"V2"
V2=100x20/1000
=2ml.
So add 2ml of 1M stock HCl plus 98ml of water to make 100ml of 0.02M HCL; this formula applies for any amount/concentration of solution. Hope this help you.
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How do you prepare 1N Hcl solution Dont tell in gL how much ml is required?
To prepare 1N HCl solution, you need to dilute concentrated HCl (usually 37% w/w) with water to get the desired concentration. For a 1N HCl solution, you would mix 1 part of concentrated HCl with 9 parts of water (ratio of 1:9). The volume needed would depend on the final volume required and the concentration of the concentrated HCl you start with.
What is the molarity of the solution produced when 85.6 g of hydrochloric acid HCl is dissolved in sufficient water to prepare 0.385 L of solution?
To find the molarity, first calculate the number of moles of HCl present: 85.6 g / 36.46 g/mol = 2.35 mol Then, divide the moles by the volume in liters: Molarity = 2.35 mol / 0.385 L ≈ 6.10 M. Thus, the molarity of the solution is approximately 6.10 M.
Prepare 500ml of 0.12N HCL solution?
To prepare 500ml of 0.12N HCl solution, you will need to dilute a concentrated hydrochloric acid solution. To do this, measure out 2.5ml of concentrated HCl (37% w/w) and dilute it to 500ml with distilled water. Be sure to wear appropriate safety equipment and handle the acid with caution.
What volume of stock solution would be required to make a 150 mL solution of 0.50 M HCl from a 2.00 M HCl stock solution?
To find the volume of the 2.00 M HCl stock solution needed, we use the formula M1V1 = M2V2. Rearranging for V1 gives us V1 = (M2 * V2) / M1, where M2 is the desired concentration (0.50 M), V2 is the final volume (150 mL), and M1 is the stock concentration (2.00 M). Plugging in the values gives V1 = (0.50 M * 150 mL) / 2.00 M = 37.5 mL. You will need 37.5 mL of the 2.00 M HCl stock solution to make a 150 mL solution of 0.50 M HCl.
How many moles of HCl are consumed in a reaction of 25 mL of 4 M is added to excess CaCO3?
To find the moles of HCl consumed, we first calculate the moles of HCl in 25 mL of 4 M solution: 25 mL * 4 mol/L = 100 mmol. Since CaCO3 is in excess, it reacts with all the HCl.
