Questions about the concentration of the solution after mixing, about the concentration of the solut

Relationship between concentration and density of ammonia:

- The thinner the ammonia, the denser it is.

- Dilute ammonia water has a high density.

After mixing 1mol l of ammonia (set to 2a%) with water and other mass:

- The mass fraction of the resulting dilute ammonia water is half of the original, i.e. a%.

- The amount and concentration of the ammonia substance before dilution (1mol l) = (1000 * density * 2a%) 17....Conversion formula for mass fraction to quantity concentration of substance).

- The amount of the substance of the resulting dilute ammonia = (1000 * density * a%) 17

- Comparing the above two formulas, if the densities are equal, the concentration of the following substance is half of the above.

- Since the density below is greater than that above, the amount concentration of the substance below is somewhat greater than half of the above.

Addendum: Don't be confused, and then another way.

- The quality of ammonia and water is m

- Volume of aqueous solution of primary ammonia: m q1....Q1 Density of ammonia before dilution, unit conversion to g l).

- The amount of NH3 in the original ammonia water: (M Q1) * 1mol L

- The mass of dilute ammonia water obtained after dilution: 2m

- Volume of dilute ammonia water obtained: 2 m q2....Q2 Density of ammonia after dilution, unit conversion to g l).

- The amount of NH3 in dilute ammonia water obtained after dilution: It is equal to the amount of NH3 in the original concentrated ammonia water, that is, (M Q1)*1mol L

- The amount and concentration of the substance of dilute ammonia obtained after dilution: [(m q1)*1mol l] (2m q2)=(q2 2q1)*1mol l

- Analyze the above equation, if q2=q1, the above equation is equal to 1mol l; However, due to Q2>Q1 (the more dilute the ammonia, the greater the density), the above formula is greater than 1mol l

Because the density of ammonia is less than that of water, the volume of ammonia is larger than that of mixed water after the same mass is mixed, so the mixed ammonia is more, and the amount of ammonia substance that must be obtained is greater than that.

The density of ammonia water of equal mass is not the same as the density of water of equal mass, and the volume is not the same, since the volume is different, then the concentration is definitely not equal to half of the distribution.

General rule: Stay close to the denser one.

Inference: If it is an equal mass mix, it should be 2x percent.

Now if you want to wait for the volume to mix, you have to have a little more dense one, so it's less than 2x.

Estimate: Set:

Mass fraction 15 The density of 1000ml of ammonia is .

The density of 1000ml of ammonia with a mass fraction of 45 is.

mnh3 = m total = nh3% = 495 1700 29

Less than 30

The more concentrated the ammonia, the less dense it is.

Let 1 kg of 1mol l of ammonia be mixed with 1 kg of water, the density of 1mol l of ammonia is A1 (A1 is less than A2), and the density of the mixed ammonia is A2 (A2 is greater than A1).

The amount of ammonia in raw ammonia: (1000 a1) * 1 1000 = 1 a1 (1000 a1 is the volume of 1 kg of ammonia).

Volume of mixed ammonia: (2000 a2) 1000 = 2 a2 concentration of mixed ammonia: (1 a1) (2 a2) = a2 2a1> because a2 > a1, so a2 2a1 >

The key to solving this problem is that "the more concentrated the ammonia, the less dense it is".

The density of ammonia water of 1mol l is because ammonia water is dense and small in size.

For example, it's all 1g, but the volume is not 2ml, you calculate.

Water is denser than ammonia, so it will be so.

1.Knowing the mass m of the solute, and the volume v of the solution, assuming that the molar mass of the solute is m The molar concentration of the solute is c=m mv2The amount of the substance of the solute is known n, and the volume of the solution v, and the molar concentration of the solute is c=n v 3

The molar mass of the solute is known to be the early mass fraction w, and the density of the solution p, the molar mass of the solute is m The molar concentration of the solute is C = 1000 pw m 4The mass of the solution m is known, and the volume v of the solution, the density of the solution w and the molar concentration of the solute is c = 1000 mw mv

Summary. The concentration of the new solution can be obtained by first calculating the amount of solute in the first solution according to the concentration and volume, and then by calculating the amount of solute in the second solution according to the concentration and volume, and then adding up the amount of solute and dividing by the total amount of the two solutions.

In a container, 4 litres of coolant were poured into the coolant, and then litres of purified water were added to the coolant (the water and coolant were mixed together). After that, I found that it was overflowing, and it was too late to fill it! The first liter of fusion liquid is released from the container, and then the liter of pure water is added to the container again, what is the ratio of coolant to dismantling water?

Now the situation is which is the coolant content and the water content, which is bigger and who is smaller? Is there more coolant or too much water? How much more?

First, the amount of solute in the first solution is calculated according to the concentration and volume, and then the amount of solute in the second solution is calculated according to the concentration and volume.

Hello, is this question a physics question? It's still math or chemistry.

Liquids must have something to do with chemistry and math<>

If it is a physics problem, you also need to consider the density <>

Okay, wait a minute, I'll figure it out.

The main thing is that 4 liters of coolant and water are mixed together and then released a noisy shed, this time the operation of the cooling liquid accounted for a large proportion of the liter, and the answer was definitely more than <> water

Sent to you.

Can you see it. This is done assuming that the coolant concentration is 100% and does not contain water, which is not the case if the coolant concentration is to be considered.

**If you don't receive it, tell me in time.

Finally, coolant: water = 32 (the simplest fraction is your own stroke), the coolant is more than water.

Senior Sister <> has worked hard, so how many milliliters is the concentration of the cooling liquid is about more than that of water? It is also necessary to add about how much water can be compared to the coolant close to 1:1 ratio of the source mode.

Thank you, sister.

The shield concentration of the coolant is 32. One liter of solution contains more coolant than water. Add a liter of water to rip the chain, which can be the first one.

The calculation process is sent to you.

How many milliliters is the concentration of coolant than water that you just asked? I don't quite understand what that means.

Can you put it another way<>

There are no units in the calculation process, and the fraction is not reduced to the simplest fraction, you can draw it yourself.

The mass fraction of solute in the solution obtained after equal volume mixing is greater than that of the solute obtained after equal mass mixing (i.e., half of the sum of the mass fraction of the solute in the two solutions).Such as sodium hydroxide, sodium chloride solution, etc.

The mass fraction of solutes in the solution obtained after equal volume mixing is smaller than that of water (i.e., half of the sum of the mass fraction of solutes in the two solutions).Such as ammonia, alcohol solution, etc.

v*p1*90%+v*p2*10%=(v*p1+v*p2)/p3 *w%

p1>p2>p3, discussed.

Excerpt from Li Zong, chemistry teacher Tan Yinguang for the college entrance examination.

First of all, I know what you mean by "equal volume mixing", [because if it is an equal mass mix, it must be equal to the mass fraction of the first two solutions of the mixing.]

The average value of . (Arithmetic mean.)

Although this conclusion is true for most solutions, it is better for ammonia than for acacia.

Solutions such as ethanol are the opposite. The explanation is like a stupid key: because no matter what the ratio is mixed, there is a mass fraction of the mixed solution = solute mass and solution mass and solution mass, we let the solute mass fraction of solution 1 be a%, the solute mass fraction of solution 2 is b%, when the mass of the Ming number is mixed [let each take 1 gram], the mass fraction of the mixed solution = (1 a% + 1 b%) 1 + 1) = (a% + b%) 2 [this is the arithmetic mean]; When the same volume is mixed, the calculation formula is still the same as above, at this time, the higher density of the solution is more concentrated, while ammonia, ethanol, etc. are small concentration.

Therefore, the mass fraction of the mixed solution deviates from the arithmetic mean and is slightly closer to the solute mass fraction of the denser solution, so the above conclusion is obtained.

If the concentration of the two is the same, then there is no need to ask, it belongs to the concentration before and after the invariant type, that is;

If the concentrations are not the same, the solution process is as follows:

Start by calculating the amount of substance in 20 mL of NaOH.

and the amount of 30 ml of the substance Xiangwu =;

Then divide the total volume by the total amount of matter, that is, (

Solution: For example, both are NaCl solutions, with a concentration of 20% and a bottle volume of 100mlThe volume of the other bottle is 200ml, and the concentration after mixing is found.

Answer: Concentration = mass of solute and volume of solution.

m1=100x20%=20g

m2=200x20%=40g

m=m1+m2=20g+40g=60g

v=v1+v2=100+200=300mlc%=m/v=60/300=1/5=

c%'=c%

There is no change in concentration.

Solution: For example, both are NaCl solutions, with a concentration of 20%, and a bottle of socks is 100mlThe other bottle has a volume of 200ml, and the concentration after mixing is found.

Answer: Concentration = mass of solute and volume of solution.

m1=100x20%=20g

m2=200x20%=40g

m=m1+m2=20g+40g=60g

v=v1+v2=100+200=300mlc%=m v=60 300=1 5=

c%'=c%

There is no change in concentration.

Such as hydrochloric acid and sodium hydroxide.

According to the amount of the substance 1:1 reaction slippery, and hydrochloric acid and barium hydroxide according to the amount of the substance 2:1 reaction, the same concentration, the same volume of strong acid.

And strong alkali means that the slip is the same amount of strong acid and strong alkali substance, but not necessarily completely neutralized, and the solution pH is not necessarily = 7

100 ml of 1mol L of Ca(OH)2 is mixed with 100 ml of 1mol L of HCl, oh-for, H+ is the only relinquish, oh-more, so pH > 7.

100 ml of 1mol L of NaOH is mixed with 100 ml of 1 mol L of H2SO4, oh- is, H+ is, H+ more, so the solution pH < 7.

One. The reason for the high concentration of sail removal solution.

1 day Bi Shu Ping weight possession of other substances or travel hail has rusted 2 with a graduated cylinder.

Look up at the reading as you measure the liquid.

3 When weighing solutes containing crystalline water, the solutes have weathered.

4 Look down at the tick marks when fixing volume.

5 When dissolving solid solutes or diluting liquids, transfer them to volumetric flasks before cooling to room temperature.

, volume.

Two. The reason for the low level.

1 No washing beakers.

Inner wall 2 Looking up at the volumetric flask graduation mark.

3 The solute contains other impurities.

4 The solute has deliquescent.

5 When measuring the liquid solute, look down at the reading.

6. After shaking, add water.

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