450g of sugar water with an existing concentration of 20, add 50g of sugar, what is the concentratio

450*20%=90g (this is the original solute mass) After adding 50g of solute, the total weight of the solute = 140g The mass of the solution = 450g + 50g = 500g The solute mass fraction (i.e. concentration) = 28

Solution: The dispersion of one or several substances into another substance, forming a homogeneity.

1. Stable mixture. (such as the sugar water in this question).

Solubility: A substance that is dissolved. Substances with less mass in the dissolution process are generally treated as solutes. (e.g. sugar in this question).

Solvent: A substance that dissolves other substances. Generally, substances with more mass in the dissolution process are used as solvents. But when there is water in the solution, water should be considered as a solvent.

If there is anything you don't understand, don't worry, you will learn chemistry in the third year of junior high school.

The original 450g sugar water has 450 * 20% = 90g, after adding 50g, a total of 90 + 50 = 140g of sugar, the total amount of sugar water is 450 + 50 = 500g, and the concentration is 140 500 = 28%.

The existing concentration of 20% sugar water is 450g, the sugar content is 450*20%=90g, and the water = 450-90=360g.

After adding 50g of sugar, the concentration of sugar water is: (90+50) (90+50+360)=

Summary. Suppose 20% sugar water is added xg, (30%x120+10%x) (120+x)=20% (mass divided by total mass), and x=120 is obtained

In 120g of sugar water with a concentration of 30%, how many grams of sugar water with a concentration of 10% can be added to obtain sugar water with a concentration of 20%.

Set 20% sugar water xg into the stuffy block, (30% x 120 + 10% x) (120 + x) = 20% (mass divided by total mass and quantity), and the solution is x = 120

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There are 100 grams of sugar water with a concentration of 10%, how much sugar is added to the sugar water, which is 20% stronger?

100 (1-10%) (1-20%)-100g.

Add 60 grams of sugar.

Sugar mass: 300*10%=30 grams.

The quality of the water is noisy: 300-30 = 270 grams of the world's errands.

Mass of sugar water after sugaring: 270 (Gyeongmeng 1-25%) = 360 grams.

Mass of added sugar: 360-300 = 60 grams.

It turns out that Liang Xishui accounts for 1-20% = 80%.

The content of water is 600 80% = 480 grams.

Later it was still 480 grams.

Water station 1-40% = 60%.

So now the sugar water is 480 60% = 800 grams.

Answer: 200 grams of sugar need to be added to the residue.

The existing 300 grams of sugar water with a concentration of 20% need to be turned into candy with a concentration of 40%, how many grams of sugar need to be added?

Sugar water --- candy, there is difficulty!

The concentration is equal to the solute mass to the solution mass. Grinding Meng Jing...

Add x grams of sugar.

The original sugar is grams.

Then, the added sugar is 60+x grams.

The mass of the added sugar water is 300+x grams.

So there is (60+x) (300+x)=

So x = 100 grams.

Method 1: Solution: Sugar x kg is required.

20％×40＋x）/(40+x)=40%

Solution, gotcha. x＝40/3

40 3 kg 40000 3 g.

Method 2: Solution: The pre-transport Huishu beam needs to add x kg of sugar.

x Bi Xian 40 3

40 3 kg 40000 3 g.

426 and 2 Potato empty 3 - 400

26 and 2 3 grams.

Set: A gram of sugar needs to be added.

Mass of sugar in 300g: 300g x 20% = 60g according to: solute solution = concentration.

60g+a)/(300g+a)=40%

a=100g