Chemistry

# How do you prepare a solution of 1 M H2NO4?

It depends on how much you are trying to make and what you are diluting it from, but the formula for figuring it out is Molarity of starting solution times X (in which X is how much you will be adding) equals Molarity of the solution you are producing times its volume

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## Related Questions

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 a .1 M solution, the ratio would be .1 moles of HCl per liter of water. This is equal to 3.65 grams HCl.

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) ---&gt; V1 = (1 M) x (1 litre)/(6 M) = 0.167 L or 167 ml

10N HCl is equivalent to 10 M HCl and the solution will contain 10 moles of HCl per 1 liter or solution. Depending on the volume of solution, the preparation will vary. To prepare 1 liter, place 360.5 g of HCl in a total volume of 1 L. To prepare 1 L from concentrate HCl (12.1 M) dilute 833 ml of conc. HCl to 1 L.

Magnesium sulfate is not so soluble in water to prepare a 4 M solution.

For making 1 MSolution, 40 gm NaoH dissolved in 1000 ml Distilled water .then10 ml solution of 1 M (Stock solution) dissolved in 500 ml distilled water for 0.02 M NaOH solution. R.C Joshi.

Dissolve 74.55g of KCL in some water then volume it up to 1 liter.

use the equation that is standard: 1000 ml 1 M solution= (MOLECULAR WEIGHT) X ml 0.05 M solution = ((MOLECULAR WEIGHT)*X*0.05)/1000

impossible....u want to use a low contrated solution to dilute into a higher contration solution? No way man,...

one mole of HCl in one litre of water, result in 1M HCl

Prepare HCl 1 M by HCl concentration 37 % HCl concentration 37 % have density =1.19 g/ml HCl 1 M use HCl 37 % 82.81 ml make volume with water to 1 liter

how will you prepare 0.38M sodium acetate solution

Exactly how you prepare will depend on what you are starting with. Typically to make a 1 M HCl solution, you will be starting with a stock solution of more concentrated HCl that you will then dilute. See the Related Questions for complete instructions on how to prepare a solution by diluting a stock solution. 1M HCl Solution*Concentrated HCl. its 37.5%. *The density of concentrated HCl is 1.189g/ml we will need this number as well, and of course the atomic mass of HCl 36.46 * Calculation=((37.5/100)(1000)(1.189))/36.46 = 12.2M *If HCl concentrated is 12.2 M then to make a 1L solution of 1M HCL (12.2) x = 1(1) x = 1(1)/(12.2) x=0.082ml/ml of water x = 82mL HCl per liter

Concentration of NaOH = 0.025 M = 0.025 Moles per Litre of SolutionVolume of Solution required = 5.00LWe can say therefore that:Number of Moles of NaOH needed to prepare the solution= Concentration of NaOH * Volume of Solution requiredTherefore:Number of Moles of NaOH needed to prepare the solution= 0.025M * 5.00L= 0.125molesFrom this we can say that 0.125 moles of NaOH are needed to prepare a 5.00 L solution with a concentration of 0.025M of NaOH.

To prepare the buffer using solid form reagents, prepare a 0.1 M ammonium acetate solution by dissolving 7.7 g ammonium acetate in a 1000 ml water. Adjust 1 L of this solution to pH 4.5 by adding acetic acid (about 8 ml) and 5 ml of 1 M p-TSA (equivalent to 5 mM p-TSA).

You prepare a solution by dissolving a known mass of solute into a specific amount of solvent. In solutions, M is the molarity, or moles of solute per liter of solution. For 300 ml of a 0.1 M Na CL solution from a solid Na CL solution and water you need water and sodium chloride.

First, you will need about 5.7g of KOH. Then you will dissolve the KOH solution in 1dm of water.

8.3ml HCl from 37% v/v stock bottle is required to prepare 0.1 M HCl solution.(answered by Prof.WASEEM UR RAHMAN KHAN)

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