Solutions of percentage concentration. For cooking we need

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 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 cases where it is necessary to prepare 1 liter of a 15% sodium chloride solution, required amount salts are 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.

I've always believed that pizza requires special ovens and special dough. When I watched how they make it in restaurants, I saw a lot of professional equipment and did not even think that it was possible to make such a dough at home. After all, the dough is the basis on which it depends whether you succeed real pizza or not.

It turns out you can. And there are so many ways to make real dough at home.

In our family, love for this dish is divided into two camps: children love the dough to be more magnificent, and my wife and I believe that the main thing is the filling, and the dough should not be the main focus.

If you share an adult point of view, then let me tell you how to make dough without yeast at home so that it turns out the same as in a pizzeria.

Pizza dough without yeast on water - 5 minute recipe

Let's start with my favorite and probably the most simple recipe. If you have sausage, tomatoes and cheese in the refrigerator, you want something tasty, but you don’t want to go to the store, then this method of making dough will help you out.

It is done quickly, literally in 5-10 minutes, depending on your culinary skills.

Ingredients:

• 2 cups of flour
• 1 egg
• 200 ml water
• 20 ml refined sunflower oil
• Sugar - 1 tsp
• Salt - 1 tsp
• Baking powder - 1 tsp

If you are annoyed that the amount is given in glasses and you do not understand how much it is in grams, because. glasses are all different, then I’ll tell you a little secret - the Soviet faceted glass has historically been taken as the standard for the “glass” measure. Its capacity is 200 ml (to the rim) or 250 ml (if filled to the brim)

Cooking:

1. Break an egg into a bowl. Add salt and sugar and mix thoroughly.

2. Add to the bowl vegetable oil.

The oil must be refined so that there is no specific smell.

3. And a glass of water. Mix until smooth.

The water should be warm, but not hot. One that you can dip your finger into.

4. Cooking the dough.

Add a teaspoon of baking powder to the flour. We mix a little.

Slowly pour the previously prepared mixture into the flour, mixing along the way.

5. Now we carefully knead the flour, knead and roll it out with our hands until a homogeneous piece of dough is obtained.

The final amount of dough is enough for you to prepare three bases with a diameter of 27 cm (this is the diameter of a standard frying pan).

If you want to cook only one pizza, then divide the dough into 3 parts, take one part for cooking, and put the other two in a bag in the freezer until the next cooking. In frozen form, you can store the dough as long as you like without loss of taste properties.

6. We take the remaining piece of dough and roll it out with a rolling pin to a thickness of 3-5 millimeters.

7. After that, it remains only to put it on a greased pan and give it the shape of this same pan with your hands (so that the filling does not spread).

In no case should the pan be cast iron. You should try to find a pan with the thinnest walls, such as pancake pans. This is necessary so that the dough has time to bake, and the cheese does not have time to dry out.

The base is ready, put your favorite foods in it and send it to the oven.

Baking rules:

• The readiness of the pizza is determined by the readiness of the cheese. Once it melts and gurgles, it's ready. This usually takes about 10 minutes at the highest oven temperature. In this case, the pan must be placed on the lowest shelf (in a gas oven).
• If there are any products in the filling that require heat treatment(onions, chicken, meat), then they must first be fried, because even onions will not have time to cook in 10 minutes of baking.
• Do not forget that before putting the ingredients on the dough, it must be greased. tomato paste or ketchup. You can try to cook special sauce independently, but most often, in the end, just tomato paste is obtained
• The cheese is placed twice - the first time on the sauce, the second - the final layer on top of all the ingredients. In both cases use the same number cheese.

How to cook dough with milk or kefir

If you want the dough to be more tender, you can use milk or kefir instead of water. There is not much difference between these two ingredients and their quantity in the recipe is the same.

Ingredients:

• Kefir (milk) - 500 ml
• Wheat flour - 650 g
• Eggs - 2 pcs
• Soda - 0.5 tsp
• Salt - 0.5 tsp
• Refined sunflower oil - 2 tbsp

Cooking:

1. Pour kefir (or milk) into a bowl, break eggs into the same place and add vegetable oil. Beat the mixture with a whisk.

2. Add soda and salt to the mixture and beat well again.

3. Add a third of the flour.

It is not necessary to fall asleep all the flour at once, otherwise it will be difficult to get rid of the lumps formed.

4. We begin to stir the dough with a spoon, gradually adding flour. When the dough turns into one lump, continue to knead it and knead it with your hands.

5. You need to mix the dough until it stops sticking.

After that, lightly sprinkle the dough with flour. It does not require proofing and you can immediately start cooking pizza.

Recipe for yeast-free dough without eggs on the water

Quite simple, but no less delicious option in case there was no kefir or eggs in the refrigerator, but you still want pizza.

Ingredients:

• Wheat flour - 1 cup (or 230 gr)
• Baking powder - 1 tsp
• Salt - 0.5 tsp
• Water - 100 ml
• Olive oil - 4 tbsp (can be sunflower)

Cooking:

1. Add salt and baking powder (or soda) to a bowl of flour and mix with a spoon.

2. Make a small indentation in the flour and pour warm water and vegetable oil.

3. Mix all the ingredients. First, do this with a spoon, and then knead the dough with your hands.

4. As a result, we get a very soft, plastic and pliable mass. Before you start rolling out the dough, let it “rest” for about 20 minutes and start preparing the filling. During this time, the baking powder in the dough will begin to work and it will turn out soft and airy.

Pizza dough without yeast step by step video recipe

Well, for those who are interested in the nuances of cooking and option finished stuffing I suggest to look detailed video a recipe that reveals a lot of interesting things.

In the following articles, we will pay more attention to the filling and its correct selection and combination, so that it is tasty and healthy.

Thank you for your attention.

The weight percentage indicates what percentage of the total weight of the solution is the solute.

■ 18. How much potassium nitrate should be taken to prepare 300 g of 2% salt solution?
19? How much water and sugar is required to prepare 250 g of a 10% solution?
20. How much barium chloride is required to prepare 50 g of a 0.5% solution?

In laboratory practice, one often has to deal with crystalline hydrates - salts containing crystallization water, for example CuSO 4 5H 2 O, FeSO 4 7H 2 O, etc. In this case, one should be able to take into account crystallization water.

Example 2 How much crystalline hydrate blue vitriol must be weighed to obtain 200 g of a 5% solution of copper sulfate? How much water do you need to take for this?
Given: 200 g 5% CuSO 4	Solution: First you need to determine how much copper sulfate CuSO4 is needed to prepare a given amount of solution: (200 5): 100 \u003d 10 g CuSO 4. 160 g CuSO 4 - in 250 g CuSO 4 5H 2 O 10 "CuSO 4 -" x "CuSO 4 5H 2 O X=(250 10) : 160 = 15.625 g Water is required to prepare the solution 200- 16.6 15= 184.375 g.
CuSO 4 5H 2 O (g)?

■ 21. How much Na 2 SO 4 10H 2 O crystalline hydrate will be needed to prepare 2 kg 34 Na 2 SO 4 solution?
22. How much iron sulfate FeSO4 7H2O will be required to prepare 30 kg of a 0.5% FeSO 4 solution?
23. How much CaCl 2 6H 2 O crystalline hydrate will be required to prepare 500 g of 10% CaCl 2 solution?
24. How much ZnSO 4 7H 2 O crystalline hydrate will be required to prepare 400 g of a 0.1% ZuSO 4 solution?

Sometimes it is necessary to prepare solutions of a certain percentage concentration, using other, more concentrated solutions for this. This is especially often encountered in the laboratory when obtaining solutions of acids of different concentrations.

Example 3 How much 80% sulfuric acid is required to prepare 200 g of a 10% solution of this acid? We denote the mass of the first solution m 1, the mass of the second - m 2, the concentration of the first solution C 1, the concentration of the second solution C 2.
Given: m 1 = 200g C1 = 10% C2 = 80%	First of all, you need to find out how much pure anhydrous chamois acid will be required to prepare 200 g of a 10% solution: (200 10): 100 = 20 g. We determine how much 80% sulfuric acid contains 20 g of pure acid, arguing as follows: in 100 g 80% H 2 SO 4 - 80 g pure H 2 SO 4 "x" 80% H 2 SO 4 -20 "" H 2 SO 4. Hence x \u003d (100 20): 80 \u003d 25 g of an 80% solution. Therefore, for our purpose, we need 25 g of an 80% solution of H 2 SO 4 and 200-25 \u003d 175 g of water.
m2 (g) ?

■ 25. How much 80% phosphoric acid is required to prepare 2 kg of a 5% solution?
26. How much 20% alkali is required to prepare 5 kg. 1% solution?
27. How much 15% nitric acid is required to prepare 700 g of a 5% solution?
28. How much 40% sulfuric acid is required to prepare 4 kg of a 2% solution?
29. How much 10% hydrochloric acid is required to prepare 500 g of a 0.5% solution?

However, to make the correct calculation is not all for the laboratory assistant. You need to be able not only to calculate, but also to prepare an acid solution. But acids cannot be weighed on a scale, they can only be measured using measuring utensils. Measuring utensils are designed to measure volume, not weight. Therefore, you need to be able to calculate the volume of the found amount of solution. It can't be done without knowing specific gravity(density) of the solution.
Let us turn again to example 3, given on page 67. From the table (Annex III, paragraph 3, page 394) it can be seen that 80% has a density d\u003d 1.7, and the mass of the solution R\u003d 25 g. Therefore, according to the formula

V \u003d P: d we find: V \u003d 25: 1.7 \u003d 14.7 ml.

The density of water is practically considered equal to unity. Therefore, 175 g of water will occupy a volume of 175 ml. Thus, to prepare 200 g of a 10% solution of 80% sulfuric acid, you should take 175 ml of water and pour 14.7 ml of 80% sulfuric acid into it. Mixing can be done in any chemical glassware.

■ 30. How many milliliters of 50% sulfuric acid should be taken to prepare 2 kg of a 10% solution of this acid?
31. How many milliliters of 40% sulfuric acid should be taken to prepare 5 liters of 4% sulfuric acid?
32. How many milliliters of 34% caustic potash will be required to prepare 10 liters of a 10% solution?
33. How many milliliters of 30% hydrochloric acid will be required to prepare 500 ml of 2% hydrochloric acid?

The examples of calculations that we have analyzed so far have been devoted to determining the weight or volume of the solution, as well as the amount contained in it. However, there are times when you need to determine the concentration of the solution. Let's consider the simplest case.

■ 34 25 g of salt and 35 g of water are mixed. What is the percentage concentration of the solution?

35. Mixed 5 g of acid and 75 g of water. What is the percentage concentration of the solution?

Quite often it is necessary to dilute, evaporate and mix solutions, and then determine their concentration.

■ 36. 500 g of water was added to 2 kg of a 20% solution. What was the concentration of the solution?
37. 1 liter of water was added to 5 a 36% hydrochloric acid. What was the concentration of the solution?
38. Mixed 40 kg of 2% and 10 kg of 3% solutions of the same substance. What was the concentration of the resulting solution?
39. Mixed 4 liters of 28% sulfuric acid and 500 ml of 60% sulfuric acid. What is the concentration of the resulting solution?
40. 3 kg of 20% sodium hydroxide solution was evaporated to 2 kg. What is the concentration of the resulting solution?
41. How much water should be added to 500 ml of a 30% solution (density 1.224 g / cm 3) to get a 5% solution?

To determine in what ratio the solutions should be mixed different concentrations to obtain a solution of the desired concentration, you can apply the so-called "mixing rule", or "diagonal
scheme"

■ 42. Calculate according to the diagonal scheme in what ratio the solutions should be mixed:
a) 20% and 3% to get 10%;
b) 70% and 17% to get 25%;
c) 25% and water to get 6%

Volumetric concentration of solutions. Molar concentration

When determining volumetric concentration solutions calculations are made in relation to 1 liter of solution. Molar concentration, for example, shows how many gram-molecules (moles) of a solute are contained in 1 liter of solution.
If you don't remember what a gram molecule is, refer to the appendix on page 374.
For example, if 1 liter of a solution contains 1 mol of a substance, such a solution is called one-molar (1 M), if 2 mol, two-molar (2 M), if 0.1 mol, then the solution is decimolar (0.1 M), if 0.01 mole, then the solution is centomolar (0.01 M), etc. For the preparation of solutions molar concentration you need to know the formula of the substance.

Example 7 How much sodium hydroxide do you need to take to prepare 200 ml of 0.1 M sodium hydroxide solution NaOH.
Given: V = 200 ml C = 0.1 M	Solution: Before; in total, we calculate the weight of a gram-molecule of sodium hydroxide NaOH. 23 + 16 + 1 = 40 g. Since the solution is 0.1 M, then 1 liter of the solution contains 0.1 gram-molecules of NaOH, i.e. 4 g, and 200 ml or 0.2 liters of the solution will contain an unknown amount of NaOH. We make a proportion: in 1 l of 0.1 M solution - 4 g of NaOH "0.2" "-x" NaOH From here 1:0.2 = 4:x x \u003d (4 0.2) : 1 \u003d 0.8 g. i.e., to prepare 200 ml of a 0.1 M solution, 0.8 g of NaOH is needed.
m NaOH (g)?

The molar concentration is very convenient in that equal volumes of solutions with the same molarity contain the same number of molecules, since the gram-molecule of any substance contains the same number of molecules.
Prepare a solution of molar concentration in volumetric flasks of a certain volume. On the neck of such a flask there is a mark that accurately limits the desired volume, and the inscription on the flask indicates what volume this volumetric flask is designed for.

■ 43. Calculate how much substance is required to prepare the following solutions:
a) 5 l of 0.1 M sulfuric acid solution;
b) 20 ml of 2 M hydrochloric acid solution;
c) 500 ml of 0.25 M aluminum sulfate solution;
d) 250 ml of 0.5 M calcium chloride solution.
Solutions of acids of molar concentration often have to be prepared from percentage solutions.

■ 44. How much 50% nitric acid is required to prepare 500 ml of a 0.5 M solution.
45. What volume of 98% sulfuric acid is needed to prepare 10 liters of 3 M solution?
46. ​​Calculate the molarity of the following solutions:
a) 20% sulfuric acid;
b) 4% sodium hydroxide;
c) 10% nitric acid;
d) 50% caustic potash.

Normal concentration of solutions

Normal is expressed as the number of gram equivalents of a solute in 1 liter of solution. In order to make a calculation for the Preparation of a solution of normal concentration, you need to know what an equivalent is. The word "equivalent" means "equivalent".
The equivalent is the weight amount of an element that can combine with 1 weight part of hydrogen or replace it in compounds.
If a water molecule H 2 O contains two hydrogen atoms, weighing a total of 2 y. e., and one oxygen atom weighing 16 c.u. e., then 1 u. e. hydrogen accounts for 8 y. e. oxygen, which will be the equivalent of oxygen. If we take some oxide, for example ferrous oxide FeO, then there is no hydrogen in it, but there is, and we found from the previous calculation that 8 y. units of oxygen are equivalent to 1 y. e. hydrogen. Therefore, it is enough to find the amount of iron that can combine with 8 cu. e. oxygen, and this will also be its equivalent. The atomic weight of iron is 56. In oxide, 56 cu. e. Fe accounts for 16 y. e. oxygen, and at 8 y. That is, iron oxygen will have to be half as much.
You can find an equivalent for complex substances, for example, for sulfuric acid H 2 SO 4. In sulfuric acid for 1 u. i.e. hydrogen accounts for half of the acid molecule (including, of course, and), since the acid is dibasic, i.e., the equivalent of sulfuric acid is equal to its molecular weight (98 c.u.) divided by 2, i.e. 49 c.u. e.
The equivalent for a base can be found by dividing it by metal. For example, the equivalent of NaOH is equal to the molecular weight (40 cu) divided by 1, i.e. sodium. The NaOH equivalent is 40 cu. e. The calcium equivalent of Ca (OH) 2 is equal to the molecular weight (74 cu) divided by calcium, namely 2, i.e. 37 y, e.
In order to find an equivalent for any salt, you need to divide it by the valency of the metal and the number of its atoms. So, aluminum sulfate Al 2 (SO 4) 3 is equal to 342 cu. e. Its equivalent is: 342: (3 2) = 57 c.u. where 3 is the valency of aluminum and 2 is the number of aluminum atoms.
■ 47. Calculate the equivalents of the following compounds; a) phosphoric acid; b) barium; c) sodium sulfate; d) aluminum nitrate.

The gram equivalent is the number of grams of a substance that is numerically equal to the equivalent.
If 1 liter of solution contains 1 gram equivalent (g-equivalent) of a solute, then the solution is one-normal (1 N), if 0.1 gram-equivalent, then deci-normal (0.1 N), if 0.01 gram equivalent, then centinormal (0.01 n.), etc. To calculate the normal concentration of solutions, you also need to know the formula of the substance.

Solutions of normal concentration, like molar solutions, are prepared in volumetric flasks.
■ 48. How much sulfuric acid is needed to prepare 2 liters of 0.1 N. solution?
49. How much aluminum nitrate should be taken to prepare 200 ml of 0.5 N. solution?
It is often necessary to prepare solutions of normal concentration from concentrated solutions of percentage concentration. This is done in the same way as when preparing solutions of molar concentration, but not the gram-molecular, but the gram-equivalent weight is calculated.

How much 60% nitric acid should be taken to prepare 200 ml of 3N. solution? 61. What volume of 20% sulfuric acid is needed to prepare 20 liters of 0.1 N. solution?
Normal concentration is very convenient, since if the solutions have the same normality, then equal volumes of these solutions contain equivalent amounts of solutes. Therefore, substances contained in equal volumes of such solutions react completely (if, of course, they can react with each other at all). For example, if you take 1.5 liters of 0.1 n. sodium hydroxide solution and 0.1 N. hydrochloric acid solution, then 1.5 liters of hydrochloric acid solution will also be required to react with caustic soda.

In a previous article, we talked about all the recommendations that patients should follow during the therapeutic diet No. 5. In this article, we bring to your attention the most popular and delicious table recipes number 5.

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Potato soup with chicken

This excellent dish will satiate the patient and bring his body many benefits. You can from time to time change the composition of products in this soup. For example, you can add cauliflower, zucchini, broccoli, rice, pasta.

For cooking we need:

200 g chicken breast , 2-3 potatoes, small carrot, small onion and salt to taste.

• Cut the chicken into pieces and cook until cooked;
• While the chicken is cooking, peel the potatoes, cut them into small cubes, peel the carrots and onions and also cut them into cubes;
• Boil all vegetables separate saucepan until ready;
• Then we shift the chicken pieces to the vegetables and vegetable broth, and chop with a blender until uniform consistency. Soup puree is ready!

Diet pilaf

As you know, pilaf is a non-dietary dish. But you can cook it a little easier, without harmful additives. We will need the following ingredients:

• 500 g of lean beef;
• 700 g of rice;
• 4 medium carrots;
• small bulb;
• salt to taste.

Boil the beef in several waters until it becomes soft and tender. After each boil, drain the water, rinse the meat in water, and put it in clean water and bring to a boil. Instead of beef, you can cook pilaf with chicken.

Cool the boiled beef and cut into pieces. Then we prepare the vegetables. Carrots are peeled and grated coarse grater. The onion is cut into small cubes. We put them in a cauldron, add boiled meat and lightly simmer over low heat.

Don't forget to salt them. Add vegetables and meat on top thoroughly washed rice. Align it and fill it clean water. Water should be more than rice by about 1 finger. During cooking, rice swells and absorbs a lot of water. Put the cauldron on slow fire, and stir the contents from time to time. Cook the rice until it has absorbed all the water, adding more hot water if necessary. boiled water. At the end, mix well and season with salt to taste.

Cabbage rolls for diet No. 5

Recipe diet cabbage rolls You can pamper not only adults, but also small family members. We will need:

• Head of cabbage;
• 1-2 small carrots;
• 2 tomatoes;
• Small bulb;
• boiled egg;
• A glass of rice;
• vegetable broth;
• Salt.

The cabbage is boiled whole until it becomes soft. Then it splits into individual leaves. Boil rice in a separate saucepan until tender. The boiled egg is cut into cubes. We cut the onion into cubes, three carrots on a coarse grater, three tomatoes on a grater, then stew them a little in a pan. Rice mix with egg, stewed vegetables. The resulting stuffing is wrapped in cabbage leaves, put in a cauldron, pour vegetable broth and simmer until done. When serving, you can decorate with herbs and low-fat sour cream.

Noodles with meat

For this recipe you will need:

• 500 g pork without fat;
• 2-3 bulbs;
• 300 g spaghetti;
• 1 bell pepper;
• 5 chicken eggs;
• 100 g of mild and low-fat cheese;
• A little low fat milk or sour cream;
• salt, herbs, seasonings.

The meat is passed through a meat grinder and crushed to mince. The onion is cut into cubes and stewed a small amount sunflower oil and water. Then minced meat is added to the onion and stewed until tender. bell pepper should be cut into strips, and boil the spaghetti until tender in salted water. Cheese grate on a medium-sized grater. Mix spaghetti with pepper, and put in a prepared baking dish. A layer is laid on top minced meat with onion. Sprinkle with grated cheese. Prepare the omelette mass for pouring: beat eggs with milk or sour cream. egg mixture water the noodles. Then put it in the oven for about half an hour at a temperature of 200 ° C until golden brown. Ready meal cut into pieces.

lazy dumplings

This dish will also appeal to both children and adults. Required Ingredients:

• 0.5 kg of cottage cheese;
• 2 tbsp. spoons of sugar;
• 1 egg;
• 150 g flour;
• a pinch of salt.

Add to curd a raw egg, a pinch of salt and mix thoroughly. Then add the required amount of sugar. Then add flour to the mixture and mix until soft dough. Ready dough should be slightly damp and slightly sticky to the hands. To make it easier to work with the test, wet your hands with water. Roll a small “sausage” out of a piece of dough, cut it into equal pieces. Give them the desired shape. You can cook some dumplings right away, you can freeze the other part. Boil dumplings for about 2-3 minutes until they float to the surface. So you can always cook tasty and hearty meal on hastily for their households.

Rice casserole

Sweet rice casserole- very tasty and airy dessert. For cooking we need:

• A glass of rice;
• 200 g of cottage cheese;
• 3 chicken eggs;
• 3 small sweet apples;
• a handful of raisins;
• 2 tablespoons of sugar;
• 2 glasses of milk;
• 1 tablespoon sour cream.

Add a glass of water to the milk, and boil the rice in milk until tender. Grind the cottage cheese through a sieve so that it is more gentle and airy. Whisk eggs with sugar. Rinse the apples, remove the peel and seeds, and cut into cubes. Drain the liquid from the rice and let it cool slightly. Then add cottage cheese, raisins, diced apples, eggs with sugar to it. Mix the resulting mass thoroughly and put in a baking dish. Beat 1 egg with a spoonful of sour cream and pour the casserole with the resulting mixture. Bake until cooked through in an oven preheated to 200 degrees about 20-25 minutes.

Berry kissel

This drink is very tasty, helps to make more varied menu. Used warm. Cold drinks during diet number 5 prohibited. Required Ingredients:

• sweet berries: raspberries, strawberries, blueberries, cherries, etc.
• 2 tablespoons of starch;
• 4 tablespoons of sugar;
• 2 liters of pure water.

Rinse the berries thoroughly, if necessary, remove the stones. Dissolve starch in a small amount of water. Pour berries and sugar with water and cook until boiling, then pour in the diluted starch, and cook, stirring, for 15-20 minutes to thicken. Kissel is ready.

Rosehip decoction

To prepare this medicinal decoction you will need:

• rose hip;
• water.

Proportion of wild rose to water when preparing a decoction is 1:10 . That is, water is needed 10 times more than berries. To prepare a decoction, rose hips should be crushed, then the decoction will be less infused. Rosehip is filled with water and brought to a boil. Then the broth is removed from the fire, covered with a lid and infused for 12 hours. You can add some sugar if you wish.

Video recipe "Carrot-apple soufflé" for diet No. 5:

(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. Ampicillin vial contains 0.5 dry medicinal product. 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

(G) 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 should be taken ( dry matter) in g and water to prepare 1 liter of 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. For setting a warm compress from a 40% solution 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.