Preparation and use of soda solution. Calculations in the preparation of solutions of percentage concentration

To the question, tell me how to make a 1% solution with copper sulfate and lime (or soda ash) given by the author Larysa lymar the best answer is Bordeaux liquid is not prepared with soda ash. Soda ash is sodium carbonate, and you need calcium hydroxide (slaked lime). Otherwise, you will get malachite. How to cook - read:

Answer from Mujtahid.[guru]

Answer from spacer[guru]

Answer from I-beam[guru]

Answer from Special[guru]
to prepare a 1% copper solution, dilute 100 g of vitriol in 10 liters of water and neutralize 100 g of slaked lime ==>>

Answer from Natali Natali[guru]

Answer from Zhanna S[guru]


1-% BORDEAUX mixture.



and also diluted with 5 liters of water.















(on green leaves).
1-% BURGUNDE liquid


Add 50 g. soap.
COPPER_SOAP SOLUTION.















+ Soap in the same amount.
Success in the fight against diseases)

Answer from Mujtahid.[guru]
Buy better ready-made Bordeaux liquid.

Answer from Galina Russova (Churkina) GALJ[guru]
PROCESS PLANTS FROM PHYTOFTORA OF OTHER DISEASES

Answer from Elena Akentieva[guru]
Do not suffer with the Bordeaux mixture, a very inconvenient preparation in terms of preparation (does not mix well) and processing (clogs the sprayer). Buy Ordan or Abiga-Peak, wonderful fungicides, no hassle.

Answer from Kostenko Sergey[guru]
to prepare a 1% copper solution, dilute 100 g of vitriol in 10 liters of water and neutralize 100 g of slaked lime ==>> 10 liters of Bordeaux with a copper concentration of 1%

Answer from Natali Natali[guru]
Dilute 100 g of copper sulfate in 5 liters of warm water and separately dilute 100 g of lime in 5 liters of water. Then pour the solution of vitriol into the solution of lime - NOT vice versa and get a 1% solution of Bordeaux liquid. In other words: for 10 liters of water - 100 g of vitriol and lime

Answer from Zhanna S[guru]
Bordeaux mixture is prepared on the basis of copper sulphate and lime.
BURGUNDY - from copper sulphate and soda (food can also be used) + soap.
1-% BORDEAUX mixture.
100 g of quicklime and quenched in a small amount of water, diluted with water to 5 liters, milk of lime is obtained.
In another container (non-metal)
dissolve 100 g of copper sulfate in hot water
and also diluted with 5 liters of water.
A solution of copper sulphate is poured into milk of lime and mixed well.
Copper to lime, not the other way around!
You can dissolve 100 g of copper sulfate in 1 liter of hot water
and pour, gradually stirring, into 9 liters of milk of lime.
But mixing both concentrated solutions,
and then diluted with water up to 10 liters is unacceptable.
It turns out a mixture of poor quality.
Properly prepared liquid has a turquoise, sky blue color and a neutral or slightly alkaline reaction.
Acidity is tested with litmus paper.
which is dyed blue.
You can lower any clean (not rusty) piece of iron into the solution.
In an acidic environment, copper actively settles on iron.
The acidic mixture will burn the leaves.
It is neutralized by adding milk of lime.
1% mixture is applied to vegetative plants
(on green leaves).
1-% BURGUNDE liquid
100 g of copper sulfate and 100 g of soda ash per 10 liters of water. Dissolve separately.
merge together, check the acidity,
i.e., they are prepared in the same way as the Bordeaux mixture, only the lime is replaced with soda.
Add 50 g. soap.
COPPER_SOAP SOLUTION.
10 g of copper sulphate are dissolved in 0.5 l of hot water.
Separately, 100 g of soap is diluted in 10 liters of water (preferably warm).
A solution of copper sulphate is poured into a soap solution in a thin stream with constant stirring.
The drug is prepared before spraying.
Copper-soap preparation (emulsion) can be prepared in higher concentrations
(20 g of copper sulfate and 200 g of soap
or 30 g of vitriol and 300 g of soap per 10 liters of water).
A properly prepared emulsion should have a greenish color and not form flakes.
To avoid coagulation of the drug in cases of its preparation in hard water,
reduce the amount of copper sulfate
or add 0.5% (50 g per 10 l of water) soda ash (linen) to the water.
Can be used in conjunction with karbofos (20 g per 10 l of emulsion)
for simultaneous control of aphids and spider mites.
Soda ash, or laundry, soda (sodium carbonate) is a white crystalline powder, soluble in water.
Used to control powdery mildew
at a concentration of 0.5% (50 g per 10 liters of water).
+ Soap in the same amount.
To prepare a working solution, dilute soap in soft water and add soda, previously dissolved in a small amount of water.
Success in the fight against diseases)

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.

Determine what you know and what you don't. In chemistry, dilution usually means obtaining a small amount of a solution of known concentration, then diluting it with a neutral liquid (such as water) and thus obtaining a less concentrated solution of a larger volume. This operation is very often used in chemical laboratories, therefore, reagents are stored in them in a concentrated form for convenience and diluted if necessary. In practice, as a rule, you know the initial concentration, as well as the concentration and volume of the solution that you want to receive; wherein the volume of the concentrated solution to be diluted is unknown.

• In another situation, for example, when solving a school problem in chemistry, another quantity may act as an unknown: for example, you are given an initial volume and concentration, and you need to find the final concentration of the final solution with a known volume. In any case, it is useful to write down the known and unknown quantities before starting the problem.
• Consider an example. Suppose we need to dilute a solution with a concentration of 5 M in order to obtain a solution with a concentration of 1 mm. In this case, we know the concentration of the initial solution, as well as the volume and concentration of the solution to be obtained; Not the volume of the initial solution to be diluted with water is known.
  • Remember: in chemistry, M is a measure of concentration, also called molarity, which corresponds to the number of moles of a substance per 1 liter of solution.

Substitute the known values ​​into the formula C 1 V 1 = C 2 V 2 . In this formula, C 1 is the concentration of the initial solution, V 1 is its volume, C 2 is the concentration of the final solution, and V 2 is its volume. From the resulting equation, you can easily determine the desired value.

• Sometimes it's useful to put a question mark in front of the value you're looking for.
• Let's go back to our example. Substitute the known values ​​into the equation:
  • C 1 V 1 = C 2 V 2
  • (5 M)V 1 = (1 mM) (1 L). Concentrations have different units of measure. Let's take a closer look at this.

Take into account any difference in units of measure. Since dilution leads to a decrease in concentration, and often a significant one, concentrations are sometimes measured in different units. If you miss this, you can make a mistake with the result by several orders of magnitude. Convert all concentration and volume values ​​to the same units before solving the equation.

• In our case, two concentration units are used, M and mM. Let's convert everything to M:
  • 1 mM × 1 M/1.000 mM
  • = 0.001M.

Let's solve the equation. When you have reduced all the quantities to the same units of measure, you can solve the equation. To solve it, knowledge of simple algebraic operations is almost always sufficient.

• For our example: (5 M)V 1 = (1 mM) (1 L). Bringing everything to the same units, we solve the equation for V 1 .
  • (5 M)V 1 = (0.001 M) (1 L)
  • V 1 \u003d (0.001 M) (1 l) / (5 M).
  • V 1 = 0.0002 l, or 0.2 ml.

Think about applying the result in practice. Suppose you have calculated the required value, but still find it difficult to prepare a real solution. This situation is quite understandable - the language of mathematics and pure science is sometimes far from the real world. If you already know all four quantities in the equation C 1 V 1 \u003d C 2 V 2, proceed as follows:

• Measure the volume V 1 of the solution with concentration C 1 . Then add diluting liquid (water, etc.) so that the volume of the solution becomes V 2 . This new solution will have the required concentration (C 2).
• In our example, we first measure 0.2 ml of a stock solution with a concentration of 5 M. Then we dilute it with water to a volume of 1 liter: add 999.8 ml of water to it. The resulting solution will have the required concentration of 1 mM.

(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 to 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 ingredient

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% solution by volume 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.