Formula for solving problems for diluting solutions. Calculations in the preparation of aqueous solutions

(obtain a less concentrated solution from a more concentrated solution)

1 action:

Number of ml of a more concentrated solution (to be diluted)

Required volume in ml (to be prepared)

The concentration of a less concentrated solution (the one that needs to be obtained)

The concentration of a more concentrated solution (the one that we dilute)

2 action:

Number of ml of water (or diluent) = or water up to (ad) the required volume ()

Task number 6. In a vial of ampicillin is 0.5 dry drug. How much solvent should be taken in order to have 0.1 g of dry matter in 0.5 ml of solution.

Solution: when diluting the antibiotic for 0.1 g of dry powder, 0.5 ml of the solvent is taken, therefore, if,

0.1 g dry matter - 0.5 ml solvent

0.5 g of dry matter - x ml of solvent

we get:

Answer: in order to have 0.1 g of dry matter in 0.5 ml of the solution, 2.5 ml of the solvent must be taken.

Task number 7. In a vial of penicillin is 1 million units of a dry drug. How much solvent should be taken in order to have 100,000 units of dry matter in 0.5 ml of solution.

Solution: 100,000 units of dry matter - 0.5 ml of dry matter, then in 100,000 units of dry matter - 0.5 ml of dry matter.

1000000 U - x

Answer: in order to have 100,000 units of dry matter in 0.5 ml of the solution, it is necessary to take 5 ml of the solvent.

Task number 8. In a vial of oxacillin is 0.25 dry drug. How much solvent do you need to take in order to have 0.1 g of dry matter in 1 ml of solution

Solution:

1 ml of solution - 0.1 g

x ml - 0.25 g

Answer: in order to have 0.1 g of dry matter in 1 ml of the solution, 2.5 ml of the solvent must be taken.

Task #9. The price of division of an insulin syringe is 4 units. How many divisions of the syringe corresponds to 28 units. insulin? 36 units? 52 units?

Solution: In order to find out how many divisions of the syringe corresponds to 28 units. insulin needed: 28:4 = 7 (divisions).

Similarly: 36:4=9(divisions)

52:4=13(divisions)

Answer: 7, 9, 13 divisions.

Task number 10. How much you need to take a 10% solution of clarified bleach and water (in liters) to prepare 10 liters of a 5% solution.

Solution:

1) 100 g - 5g

(d) active substance

2) 100% - 10g

(ml) 10% solution

3) 10000-5000=5000 (ml) water

Answer: it is necessary to take 5000 ml of clarified bleach and 5000 ml of water.

Task number 11. How much you need to take a 10% solution of bleach and water to prepare 5 liters of a 1% solution.

Solution:

Since 100 ml contains 10 g of the active substance,

1) 100g - 1ml

5000 ml - x

(ml) active substance

2) 100% - 10ml

00 (ml) 10% solution

3) 5000-500=4500 (ml) of water.

Answer: it is necessary to take 500 ml of a 10% solution and 4500 ml of water.

Task number 12. How much you need to take a 10% solution of bleach and water to prepare 2 liters of a 0.5% solution.

Solution:

Since 100 ml contains 10 ml of the active substance,

1) 100% - 0.5 ml

0 (ml) active substance

2) 100% - 10 ml

(ml) 10% solution

3) 2000-100=1900 (ml) of water.

Answer: it is necessary to take 10 ml of a 10% solution and 1900 ml of water.

Task number 13. How much chloramine (dry matter) should be taken in g and water to prepare 1 liter of a 3% solution.

Solution:

1) 3g - 100 ml

G

2) 10000 – 300=9700ml.

Answer: to prepare 10 liters of a 3% solution, you need to take 300 g of chloramine and 9700 ml of water.

Task number 14. How much chloramine (dry) should be taken in g and water to prepare 3 liters of a 0.5% solution.

Solution:

Percentage - the amount of a substance in 100 ml.

1) 0.5 g - 100 ml

G

2) 3000 - 15 = 2985 ml.

Answer: to prepare 10 liters of a 3% solution, you need to take 15 g of chloramine and 2985 ml of water

Task number 15 . How much chloramine (dry) should be taken in g and water to prepare 5 liters of a 3% solution.

Solution:

Percentage - the amount of a substance in 100 ml.

1) 3 g - 100 ml

G

2) 5000 - 150= 4850ml.

Answer: to prepare 5 liters of a 3% solution, you need to take 150 g of chloramine and 4850 ml of water.

Task number 16. To set up a warming compress from a 40% solution of ethyl alcohol, you need to take 50 ml. How much 96% alcohol should I take to apply a warm compress?

Solution:

According to formula (1)

ml

Answer: To prepare a warming compress from a 96% solution of ethyl alcohol, you need to take 21 ml.

Task number 17. Prepare 1 liter of 1% bleach solution for inventory processing from 1 liter of stock 10% solution.

Solution: Calculate how many ml of 10% solution you need to take to prepare a 1% solution:

10g - 1000 ml

Answer: To prepare 1 liter of a 1% bleach solution, take 100 ml of a 10% solution and add 900 ml of water.

Task number 18. The patient should take the medicine 1 mg in powders 4 times a day for 7 days, then how much it is necessary to prescribe this medicine (calculation is carried out in grams).

Solution: 1g = 1000mg, therefore 1mg = 0.001g.

Calculate how much the patient needs medication per day:

4 * 0.001 g \u003d 0.004 g, therefore, for 7 days he needs:

7* 0.004 g = 0.028 g.

Answer: of this medicine, it is necessary to write out 0.028 g.

Task number 19. The patient needs to enter 400 thousand units of penicillin. Bottle of 1 million units. Dilute 1:1. How many ml of solution should be taken.

Solution: When diluted 1:1, 1 ml of the solution contains 100 thousand units of action. 1 bottle of penicillin 1 million units diluted with 10 ml of solution. If the patient needs to enter 400 thousand units, then you need to take 4 ml of the resulting solution.

Answer: you need to take 4 ml of the resulting solution.

Task number 20. Give the patient 24 units of insulin. The division price of the syringe is 0.1 ml.

Solution: 1 ml of insulin contains 40 units of insulin. 0.1 ml of insulin contains 4 units of insulin. To enter the patient 24 units of insulin, you need to take 0.6 ml of insulin.

Simple chemical solutions can be easily prepared in a variety of ways at home or at work. Whether you are making a solution from a powder material or diluting a liquid, the correct amount of each component can be easily determined. When preparing chemical solutions, remember to use personal protective equipment to avoid injury.

Steps

Calculation of percentages using the weight/volume formula

Determine the percentage by weight/volume of the solution. Percentages show how many parts of a substance are in one hundred parts of a solution. When applied to chemical solutions, this means that if the concentration is 1 percent, then 100 milliliters of the solution contains 1 gram of the substance, that is, 1 ml / 100 ml.

• For example, by weight: A 10% solution by weight contains 10 grams of the substance dissolved in 100 milliliters of the solution.
• For example, by volume: A 23% by volume solution contains 23 milliliters of the liquid compound for every 100 milliliters of solution.

1. Determine the volume of solution you want to prepare. To find out the required mass of a substance, you must first determine the final volume of the solution you need. This volume depends on how much solution you need, how often you will use it, and on the stability of the finished solution.

   • If it is necessary to use a fresh solution each time, prepare only the amount needed for one use.
   • If the solution retains its properties for a long time, you can prepare a larger amount to use it in the future.

2. Calculate the number of grams of substance required to prepare the solution. To calculate the required number of grams, use the following formula: number of grams = (required percentages)(required volume / 100 ml). In this case, the required percentages are expressed in grams, and the required volume is in milliliters.

   • Example: you need to prepare a 5% NaCl solution with a volume of 500 milliliters.
   • number of grams = (5g)(500ml/100ml) = 25 grams.
   • If NaCl is given as a solution, simply take 25 milliliters of NaCl instead of grams of powder and subtract that volume from the final volume: 25 milliliters of NaCl per 475 milliliters of water.

3. Weigh the substance. After you calculate the required mass of the substance, you should measure this amount. Take a calibrated scale, place the bowl on it and set it to zero. Weigh the required amount of the substance in grams and pour it out.

   • Before continuing to prepare the solution, be sure to clear the weighing pan of powder residue.
   • In the example above, 25 grams of NaCl needs to be weighed.

4. Dissolve the substance in the required amount of liquid. Unless otherwise specified, water is used as the solvent. Take a measuring beaker and measure out the required amount of liquid. After that, dissolve the powder material in the liquid.

   • Sign the container in which you will store the solution. Clearly indicate on it the substance and its concentration.
   • Example: Dissolve 25 grams of NaCl in 500 milliliters of water to make a 5% solution.
   • Remember that if you are diluting a liquid substance, to obtain the required amount of water, you must subtract the volume of the substance added from the final volume of the solution: 500 ml - 25 ml \u003d 475 ml of water.

   Preparation of a molecular solution

   1. Determine the molecular weight of the substance used by the formula. The formula molecular weight (or simply molecular weight) of a compound is written in grams per mole (g/mol) on the side of the bottle. If you can't find the molecular weight on the bottle, look it up online.

      • The molecular weight of a substance is the mass (in grams) of one mole of that substance.
      • Example: The molecular weight of sodium chloride (NaCl) is 58.44 g/mol.

   2. Determine the volume of the required solution in liters. It is very easy to prepare one liter of the solution, since its molarity is expressed in moles/liter, however, it may be necessary to make more or less than a liter, depending on the purpose of the solution. Use the final volume to calculate the required number of grams.

      • Example: it is necessary to prepare 50 milliliters of a solution with a molar fraction of NaCl 0.75.
      • To convert milliliters to liters, divide them by 1000 and get 0.05 liters.

   3. Calculate the number of grams needed to prepare the required molecular solution. To do this, use the following formula: number of grams = (required volume) (required molarity) (molecular weight according to the formula). Remember that the required volume is expressed in liters, the molarity is in moles per liter, and the molecular weight of the formula is in grams per mole.

      • Example: if you want to prepare 50 milliliters of a solution with a NaCl mole fraction of 0.75 (molecular weight formula: 58.44 g/mol), you should calculate the number of grams of NaCl.
      • number of grams = 0.05 L * 0.75 mol/L * 58.44 g/mol = 2.19 grams of NaCl.
      • By reducing the units of measurement, you get grams of a substance.

   4. Weigh the substance. Using a properly calibrated balance, weigh out the required amount of the substance. Place the bowl on the balance and zero before weighing. Add the substance to the bowl until you get the desired mass.

      • Clean the weighing pan after use.
      • Example: Weigh 2.19 grams of NaCl.

   5. Dissolve the powder in the required amount of liquid. Unless otherwise noted, water is used to prepare most solutions. In this case, the same volume of liquid is taken, which was used in calculating the mass of the substance. Add the substance to the water and stir it until completely dissolved.

      • Sign the container with the solution. Clearly label the solute and molarity so that the solution can be used later.
      • Example: Using a beaker (a volume measuring instrument), measure 50 milliliters of water and dissolve 2.19 grams of NaCl in it.
      • Stir the solution until the powder is completely dissolved.

   Dilution of solutions with a known concentration

   1. Determine the concentration of each solution. When diluting solutions, you need to know the concentration of the original solution and the solution that you want to receive. This method is suitable for diluting concentrated solutions.

      • Example: 75 ml of a 1.5 M NaCl solution is to be prepared from a 5 M solution. The stock solution is 5 M and needs to be diluted to 1.5 M.

   2. Determine the volume of the final solution. You need to find the volume of the solution that you want to get. You will have to calculate the amount of solution that will be required to dilute this solution to obtain the required concentration and volume.

      • Example: 75 milliliters of a 1.5 M NaCl solution are to be prepared from a 5 M initial solution. In this example, the final volume of the solution is 75 milliliters.

   3. Calculate the volume of solution that will be needed to dilute the initial solution. To do this, you will need the following formula: V 1 C 1 \u003d V 2 C 2, where V 1 is the volume of the required solution, C 1 is its concentration, V 2 is the volume of the final solution, C 2 is its concentration.

Preparation of solutions. A solution is a homogeneous mixture of two or more substances. The concentration of a solution is expressed in different ways:

in weight percent, i.e. by the number of grams of the substance contained in 100 g of the solution;

in volume percent, i.e. by the number of volume units (ml) of the substance in 100 ml of solution;

molarity, i.e. the number of gram-moles of a substance in 1 liter of solution (molar solutions);

normality, i.e. the number of gram equivalents of a solute in 1 liter of solution.

Solutions of percentage concentration. Percentage solutions are prepared as approximate, while the sample of the substance is weighed on technochemical scales, and the volumes are measured with measuring cylinders.

Several methods are used to prepare percentage solutions.

Example. It is necessary to prepare 1 kg of a 15% sodium chloride solution. How much salt is needed for this? The calculation is carried out according to the proportion:

Therefore, water for this must be taken 1000-150 \u003d 850 g.

In those cases when it is necessary to prepare 1 liter of a 15% sodium chloride solution, the required amount of salt is calculated in a different way. According to the reference book, the density of this solution is found and, multiplying it by a given volume, the mass of the required amount of solution is obtained: 1000-1.184 \u003d 1184 g.

Then follows:

Therefore, the required amount of sodium chloride is different for the preparation of 1 kg and 1 liter of solution. In cases where solutions are prepared from reagents containing crystallization water, it should be taken into account when calculating the required amount of the reagent.

Example. It is necessary to prepare 1000 ml of a 5% solution of Na2CO3 with a density of 1.050 from a salt containing water of crystallization (Na2CO3-10H2O)

The molecular weight (weight) of Na2CO3 is 106 g, the molecular weight (weight) of Na2CO3-10H2O is 286 g, from here the required amount of Na2CO3-10H2O is calculated to prepare a 5% solution:

Solutions are prepared by dilution method as follows.

Example. It is necessary to prepare 1 l of a 10% HCl solution from an acid solution with a relative density of 1.185 (37.3%). The relative density of a 10% solution is 1.047 (according to the reference table), therefore, the mass (weight) of 1 liter of such a solution is 1000X1.047 \u003d 1047 g. This amount of solution should contain pure hydrogen chloride

To determine how much 37.3% acid needs to be taken, we make up the proportion:

When preparing solutions by diluting or mixing two solutions, the diagonal scheme method or the "rule of the cross" is used to simplify calculations. At the intersection of two lines, the given concentration is written, and at both ends on the left is the concentration of the initial solutions, for the solvent it is equal to zero.

Salt solution may be needed for a variety of purposes, for example, it is part of some traditional medicine. So how to prepare a 1% solution if there are no special beakers at home to measure the amount of product? In general, even without them, you can make a 1% salt solution. How to cook it is detailed below. Before proceeding with the preparation of such a solution, you should carefully study the recipe and accurately determine the necessary ingredients. The thing is that the definition of "salt" can refer to different substances. Sometimes it turns out to be ordinary edible salt, sometimes rock or even sodium chloride. As a rule, in a detailed recipe it is always possible to find an explanation of which substance is recommended to be used. Folk recipes often also indicate magnesium sulfate, which has the second name "epsom salt".

If a substance is required, for example, for gargling or relieving pain from a tooth, then most often in this case it is recommended to use a saline solution of sodium chloride. In order for the resulting product to have healing properties and not harm the human body, only high-quality ingredients should be selected for it. For example, rock salt contains a lot of extra impurities, so instead of it it is better to use ordinary fine salt (you can also use iodized salt for rinsing). As for water, at home you should use filtered or at least boiled water. Some recipes recommend using rainwater or snow. But, given the current ecological state, this is not worth doing. Especially for residents of large cities. It is better to just thoroughly clean the tap water.

If there was no special filter at home, then the well-known "old-fashioned" method can be used to purify water. It involves freezing tap water in the freezer. As you know, in the process, it is the purest liquid that first turns into ice, and all harmful impurities and dirt sink to the bottom of the container. Without waiting for the freezing of the entire glass, you should remove the upper ice part and then melt it. Such water will be as pure and safe as possible for health. It can be used to prepare saline solution.

Now it is worth deciding on the units of measurement for liquid and solid matter. For salt, it is most convenient to use a teaspoon. As you know, it contains 7 grams of the product, if the spoon is with a slide, then 10. The latter option is more convenient to use to calculate the percentage. It is easy to measure water with an ordinary faceted glass if there are no special beakers in the house. It contains 250 milliliters of water. The mass of 250 milliliters of pure fresh water is 250 grams. It is most convenient to use half a glass of liquid or 100 grams. Next is the most difficult stage of preparing the saline solution. It is worth once again carefully studying the recipe and deciding on the proportions. If it is recommended to take a 1% salt solution in it, then in every 100 grams of liquid, 1 gram of solid will need to be dissolved. The most accurate calculations will suggest that it will be necessary to take 99 grams of water and 1 gram of salt, but such accuracy is unlikely to be required.

It is quite possible to allow some error and, for example, add one heaping teaspoon of salt to one liter of water to get a 1% saline solution. Currently, it is often used, for example, in the treatment of colds and especially sore throats. You can also add soda or a few drops of iodine to the finished solution. The resulting gargle mixture will be an excellent effective and effective remedy for sore throat. Unpleasant sensations will go away after just a few procedures. By the way, such a solution is not prohibited for use by the smallest family members. The main thing is not to overdo it with additional ingredients (especially with iodine), otherwise you can damage the oral mucosa and only aggravate the condition of a sore throat.

Also, a saline solution can be used to relieve a pulling aching toothache. True, it is more efficient to use a more saturated one, for example, 10 percent. Such a mixture is really able to relieve painful discomfort in the oral cavity for a short time. But it is not a drug, so you should never postpone a visit to the dentist after relief.

Not everyone remembers what “concentration” means and how to properly prepare a solution. If you want to get a 1% solution of any substance, then dissolve 10 g of the substance in a liter of water (or 100 g in 10 liters). Accordingly, a 2% solution contains 20 g of the substance in a liter of water (200 g in 10 liters), and so on.

If it is difficult to measure a small amount, take a larger one, prepare the so-called stock solution and then dilute it. We take 10 grams, prepare a liter of a 1% solution, pour 100 ml, bring them to a liter with water (we dilute 10 times), and a 0.1% solution is ready.

How to make a solution of copper sulfate

To prepare 10 liters of copper-soap emulsion, you need to prepare 150-200 g of soap and 9 liters of water (rain is better). Separately, 5-10 g of copper sulfate are dissolved in 1 liter of water. After that, a solution of copper sulphate is added in a thin stream to the soapy solution, while not ceasing to mix well. The result is a greenish liquid. If you mix poorly or rush, then flakes form. In this case, it is better to start the process from the very beginning.

How to prepare a 5% solution of potassium permanganate

To prepare a 5% solution, you need 5 g of potassium permanganate and 100 ml of water. First of all, pour water into the prepared container, then add the crystals. Then mix all this until a uniform and saturated purple color of the liquid. Before use, it is recommended to strain the solution through cheesecloth to remove undissolved crystals.

How to prepare a 5% urea solution

Urea is a highly concentrated nitrogen fertilizer. In this case, the granules of the substance are easily dissolved in water. To make a 5% solution, you need to take 50 g of urea and 1 liter of water or 500 g of fertilizer granules per 10 liters of water. Add granules to a container with water and mix well.