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Biology; Chemistry; Physics: Post your doubts here!

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D is for 25 cm of E,
E is for 250 cm of E,

Read the question again. They said the vinegar was taken and made to 250 cm solution by adding water. D tells us only the moles in 25 cm of the solution while E tells us moles in 250 cm of vinegar. Moles in Vinegar = Moles in 250 cm E but Moles in Vinegar != Moles in 25 of E
E tell us moles in 250 of E not of vinegar :s
 
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Pani ka poora glass lo. Us mein 100 mole hein. Usko aadha piyo ab 50 mole hein. Isi trah the difference between 250 and 25 cm of E.

Ab pani mein coloring dalo. Coloring ny aur pani to nai dala. Pani to utna hi hai bs pani mein coloring aagayi hai.

But by that pani wali example, the number of moles also should decrease 10 times from 250 to 25..
 
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Pani ka poora glass lo. Us mein 100 mole hein. Usko aadha piyo ab 50 mole hein. Isi trah the difference between 250 and 25 cm of E.

Ab pani mein coloring dalo. Coloring ny aur pani to nai dala. Pani to utna hi hai bs pani mein coloring aagayi hai.
I understand this concept but i dont understand why it can't be the same as answer to d.
D mein bhi toh they are asking for moles of acid in 25.o cm of E
and F main moles of acid in 25 of vinegar. :|
 
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I understand this concept but i dont understand why it can't be the same as answer to d.
D mein bhi toh they are asking for moles of acid in 25.o cm of E
and F main moles of acid in 25 of vinegar. :|
E is 25 cm of Vinegar + 225 cm of Water. When E decrases each material of it's composition also decreases.

But why do the moles stay the same in both 250 and 25?
In 250 the moles should be evenly spread out, and if you remove the top 25 of the volume, it should have one tenth of the original moles. Right?
What? :confused:
Gimme some maths.
 
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I understand this concept but i dont understand why it can't be the same as answer to d.
D mein bhi toh they are asking for moles of acid in 25.o cm of E
and F main moles of acid in 25 of vinegar. :|

In D, you calculated something else. That wasn't the original vinegar solution.

You added WATER to make the original solution to 250.
And then extracted 25 cm of it... so 1/10th of the moles came in the 25 cm^3 of E.

But when we are talking about 25 cm^3 of vinegar, than that would be the original solution, and it is the same as when the water was added to make it 250 cm^3 as moles remain constant. But when it gets divided, and 25 of it is taken out, then moles decrease..

Moles of Ethanoic acid in 25 cm^3 vinegar original solution= Moles of Ethanoic acid in 250 cm^3 (Since only water is added) (Now this new solution, is E, and when you take 25 out of this new E solution, that won't have the same moles as original vinegar solution)

Moles of Ethanoic acid in 25 cm^3 of E = 1/10th of moles in 250 cm^3 of E = 1/10th moles in original vinegar solution.
 
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In D, you calculated something else. That wasn't the original vinegar solution.

You added WATER to make the original solution to 250.


But when we are talking about 25 cm^3 of vinegar, than that would be the original solution, and it is the same as when the water was added to make it 250 cm^3 as moles remain constant. But when it gets divided, and 25 of it is taken out, then moles decrease..

Moles of Ethanoic acid in 25 cm^3 vinegar original solution= Moles of Ethanoic acid in 250 cm^3 (Since only water is added) (Now this new solution, is E, and when you take 25 out of this new E solution, that won't have the same moles as original vinegar solution)

Moles of Ethanoic acid in 25 cm^3 of E = 1/10th of moles in 250 cm^3 of E = 1/10th moles in original vinegar solution.
Q1 ''And then extracted 25 cm of it... so 1/10th of the moles came in the 25 cm^3 of E.''
does that mean we took only 25 cm of that 250 solution that had water+vinegar?
 
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