A junior high school chemistry problem about solubility

The first time it has been precipitated. Because of the ratio of water to solute, it is difficult to judge because it has been precipitated. So for the second time, the solution was evaporated with 60 grams of water, and after returning to 20 degrees Celsius, 6 grams of anhydrous crystals were precipitated. can be obtained.

To saturate the solution at 20 degrees Celsius, an additional 6 grams of solute need to be added.

That's it

The answer is wrong! Should a

After the previous evaporation, the solution becomes saturated.

After a saturation solution at 20 degrees Celsius, 60 grams of water were lost and 6 grams were precipitated.

The solution should be saturated at 20 degrees Celsius.

1 gram of crystals per 10 grams of water.

It must be that the answer is wrong, and the answer is the same as what the above two people said, and that's it.

The solubility of 60 and 80 is not clear, but it is still made...

Judging from the diagram, ab you also denied it, that is c

Let's take a look.

First calculate the original 40 degrees Celsius, the ammonium chloride in the ammonium chloride saturated solution is set to ammonium chloride x g

x:75=50:150 (because the formula solute:solution=s:s+100, s is the solubility) is solved to get x=25g

Then according to the formula of concentration.

c%=25g 75g+25g=25% (the definition of the concentration formula is solute solution At this time, 25g of water is added, and the solute remains unchanged as 25g, but the solution is increased by 25g, so the current mass of the solution is 100g) is obviously not 30%.

Then D is right.

If you want to know how to calculate the solubility of ammonium chloride at 60 and 80 degrees Celsius, send it to me and I can teach you how to calculate it.

The solubility at 60° and 80° is not clear.

However, according to the mass dissolved by a solid substance when it reaches saturation in 100g of solvent at a certain temperature, it is called the solubility of the substance in this solvent. It can be seen that the mass fraction of ammonium chloride in a is 40 (100+40)=, no, it is not written by you, I think you have not figured out the concept. To calculate the denominator of the mass fraction, you must remember to add the mass of the solute.

The solubility in b increases with the increase of temperature, so b is wrong.

40° in c is the mass fraction of the saturated solution is 50 (50 + 100) = 1 3, so the 75g solution contains 25g of solute, and the mass fraction after adding water is 25 (75 + 25) =, so c is wrong.

a：40/（100+40）=

B: As the temperature increases, the solubility becomes larger and becomes unsaturated.

c: original sodium chloride: 75 * (50 (50 + 100)) = 25 mass fraction after addition: 25 (75 + 25) = 25%.

d: I can't see the solubility at 80 degrees Celsius, but if d is correct, I can deduce that it is 68, right? So I can tell you the way to know that the quality of the water is constant, and there is sodium chloride at 80 degrees

84*(68 168)=34 With water: 84-34=50

Then find the saturated solution containing 50 grams of water at 60 degrees has sodium chloride: 58 2 = 29 so it all comes out, 34-29 = 5 Correct In fact, don't do this at all for a question like this Basically, you can just estimate the data at a glance and come out For you to understand better, I played so much. You should get it.

And as a junior high school student, chemistry is completely inadequate, it's a simple topic, and that's it.

The solubility is 40g, so the mass fraction is 40g (40g+100g)*100%=

b.As the temperature increases, the solubility becomes greater and the solution becomes unsaturated.

c.First, calculate the mass of the solute in the saturated solution, 50g (100g + 50g) = x 75g x = 25g

After adding water, the mass fraction was 25g (25g+75g)*100%=25%d.I can't see the solubility of the next two temperatures.

I can't read the topic clearly. However, the first question should compare the data before and after. Find a saturated solution. Find out how many grams of solute correspond to how much water you have. Good luck with it soon.

A, false. The solubility of ammonium chloride at 20 degrees is 40 grams, so the concentration of the saturated solution should be: 40 g 140g-100% =

B, false, at 20 degrees, the concentration of ammonium chloride saturated solution is, at 40 degrees, the concentration of its saturated solution is 50g 150g*100%=, so it can be seen that after heating, it will become an unsaturated solution.

c, false, at 40 degrees, the concentration of its saturated solution is, so the mass of ammonium chloride in 75g of saturated solution is 75g* = 25g, and the concentration of 25g (75g+25g) * 100% = 25% after adding 25g of water to this solution is not 30%.

To do this, ABC is wrong, D is definitely right, and the multiple-choice questions should be solved like this during the exam, quickly).

d, at 80 degrees, what is the solubility in your ** data can not be seen clearly, I will calculate it with its real solubility data to show you, and you can convert it into the data in the question** yourself. According to the data, the solubility of ammonium chloride at 80 degrees is grams, so the concentration of its saturated solution at 80 degrees = , then the mass of ammonium chloride in the saturated solution of 84 grams at 80 degrees is 84g * = (The mass of water is 84g = . Its solubility at 60 degrees is 58 grams, so the concentration of saturated solution at 60 degrees is 58g 158g*100% = , then the mass of ammonium chloride that can be dissolved at 60 degrees in gram water is set to x, then:

x (x + 100% = , get x =, that is to say, grams of water can dissolve grams of ammonium chloride at 80 degrees to get 84 grams of saturated solution, but at 60 degrees can only dissolve grams of ammonium chloride, so there will be = grams of ammonium chloride precipitated in solid form. (This data is not equal to 5 grams in option d, because the data I used to calculate was not in your **, I can't see clearly in **, you calculate it with the data in **).

You all know that A is wrong and B is wrong, the solute in C is 75* grams, and the solvent is grams, so the mass fraction is.

The most feasible is c

A False When the saturated potassium nitrate solution evaporates water, potassium nitrate crystals will be precipitated, the temperature will not change, the solubility will not change, and the mass fraction will not change.

b False Not necessarily, it is possible to evaporate a small amount of water and the solution will form a saturated solution.

c Correct After mixing: potassium nitrate mass 10g*10%+5g*40% = 3g, solution mass: 10g+5g=15g, so the mass fraction after mixing is 3g 15g*100%=20%.

d Error: Originally a%, now (10g*a%+10g) (10g+10g)*100%=2a% a=only in this case).

It is important to understand that the temperature of a is unchanged, the evaporation of the saturated solution is still a saturated solution, and the mass fraction does not change; This point will be mentioned in the teacher's lectures, pay attention to comprehend. b Evaporation, whether it reaches the saturation point with uncertainty. If it is not, the mass score will be doubled; When you reach the saturation point, it becomes a case.

This is easy to understand. c. Pay attention to the calculations. D is also as uncertain as B, because the solid is dissolved to a certain extent and does not dissolve, and the insoluble is not considered a solute, so assuming that the mass fraction is m%, the BG solid is dissolved, and 10*m% plus the dissolved B divided by 10 B.

If you dissolve it completely, it is correct.

I can't even mention this, I'm so sorry for my parents.

Question g: Because B does not precipitate, as can be seen from the figure, the content of B is less than 20 grams, that is, 40% (the solution of the solution question is all dissolved by default).

Question 2: 1) W% >A W) x 100% (because W< A+B, only the content of A should be greater than Ag).

2) w% >w - b) w x 100% (because w> a+b, only the content of b is considered to not exceed b grams).

1.It doesn't matter if it's all dissolved, as long as B is all dissolved. The mass of b in the mixture cannot be higher than 50g*20g100g=not mass fraction)

2.It's not a point.

First of all, all three solutions are saturated, A increases in solubility with temperature, and when T2 comes, A becomes an unsaturated solution, but since the mass of the solute and the mass of the solution do not change, the mass fraction of the solute does not change.

b Solubility does not change with temperature, it is always a saturated solution without any change.

The solubility of C decreases with the increase of temperature, and the solute at the temperature of T1 decreases when the solubility decreases, and the mass of the solute decreases after the solute is precipitated, and the mass fraction is smaller. But at T2, solution C is also saturated.

The answer is wrong, B is greater than A is greater than C

Because when solutions A and B reach T2 degree, the solubility of the solution increases, and there will be no crystal precipitation, while when solution C reaches T2 degree, the solubility decreases, there is crystal precipitation, and the mass fraction decreases.

Actually, I also think the answer is wrong...

My answer is b a c, and the reason is exactly what you think.

Let the original mass fraction be x and the original mass of the solution be m

The original solute was m*x

m*x)/(m/2)=26/(26+100)2x=26/126

x=13/126=

i.e. the original mass fraction is.