[allahhuakbar]

Hello. I encountered a problem stating:

"100ml of H2SO4 solution having molarity 1M and density 1.5g/ml is mixed with 400ml of water. Calculate final molarity of the H2SO4 solution, if final density is 1.25g/ml."

Here, if there is 1 mol of H2SO4 in 1000ml of solution, there will be 0.1 mol of H2SO4 in 100 ml = 9.8g of H2SO4.
Also through density, total mass of 100 ml solution is 150g.

Therefore, mass of water in 100 ml solution is 150 - 9.8 = 140.2g.

However, the volume of the water in solution is less than 100 ml (due to H2SO4 occupying some volume), and so the mass of water is less than 100g (because density of water is 1g/ml @ 4 degrees C).

Therefore, there is a contradiction in the mass of water.

Please do help. Thx

[chenbeier]

There must be a mistyping, 1 m H₂SO₄ has density 1,05 g/ cm³.
After mixing you have about 1,01 g/cm³

[Borek]

The question is perfectly solvable, you need just number of moles of the acid (don't bother converting it to mass) and final volume to find the final answer, both are pretty easy to find from the data given.

The question would be even better if whoever made it was not lazy and checked the real densities. 1 M sulfuric acid has a density of 1.0608 g/mL and 01. M solution has a density of 1.005 g/mL. Yes, the numbers given produce some confusing results, luckily they don't influence the final result.

Note that generally speaking volumes are not additive, you can't assume final volume of a mixture is a sum of volumes of its components. Masses are, so the final mass is always sum of masses of individual components.

[allahhuakbar]

All right, thanks a lot

[DrCMS]

Is the teacher being lazy or sneaky?  The density information given is incorrect but is not needed to solve the question.  People who try to use the density will get incorrect answers and maybe that is the point?

[Borek]

The density information given is incorrect but is not needed to solve the question.

The way I understand the problem - quite the opposite. You need both densities to calculate the final volume.

The longer I think about it the more I think the question is quite reasonable, I would just say "1 M solution of some substance A has a density of 1.5 g/mL".

[DrCMS]

The density information given is incorrect but is not needed to solve the question.

The way I understand the problem - quite the opposite. You need both densities to calculate the final volume.

The longer I think about it the more I think the question is quite reasonable, I would just say "1 M solution of some substance A has a density of 1.5 g/mL".

I see what you are saying.  I was ignoring the densities as not real and assuming final volume of the dilute solution was 500ml (within 1% error) but yes if this is actually "substance A" and the densities are real then they are needed to determine the correct answer. 

[MNIO]

I'm a little confused about the discussion here.  This problem is clearly
  (1) an example that volume is not conserved but mass is
  (2) has fake densities (or at least typo-ed densities)

just to be clear
  volume 1M H2SO4 + volume H2O ≠ volume mixture
  mass 1M H2SO4 + mass H2O = mass mixture
  moles H2SO4 in 1M H2SO4 = moles H2SO4 in dilute solution.. the moles don't change

steps to solve this
  (1) convert volume 1M H2SO4 to mass 1M H2SO4 using density H2SO4
  (2) convert volume H2O to mass H2O using density of H2O (not given)
  (3) add the masses to get mass solution
  (4) convert mass solution to volume solution using density of dilute solution
  (5) at this point you can either do this
          from a mole balance M1V1 = M2V2
       or
         convert 100.mL 1M H2SO4 to moles H2SO4, then divide by volume solution

*****
I'll demonstrate the math, you all can correct the calcs with the correct densities if you like


this is NOT correct
  M2 = 1M * (100mL / 500mL) = 0.20mL

One of you all can update the densities and recalc this if you wish.

[Fish200398]

If we using both density, the final volume was 440 mL. What happened to the rest 60 mL ? Because earlier we have 100 mL, then added 400 mL ?

Also known 100 mL of 1 M H2SO4
That will consist of 5.32 mL H2SO4 and 94.6 mL water.
The weight should be 9.8 g H2SO4 + 94.6 g water = 104.4 g not 150 g, the density is wrong?

[Borek]

If we using both density, the final volume was 440 mL. What happened to the rest 60 mL ?

It was stated several times in the thread: volumes are not additive. Volume of the mixture is almost never exactly sum of volumes of its components.

For diluted solutions the difference is typically negligible.

Also known 100 mL of 1 M H2SO4
That will consist of 5.32 mL H2SO4 and 94.6 mL water.

No, it won't. The only thing we can say for sure is that it will consist of 0.1 moles of H₂SO₄ and amount of water needed to fill up to 100 mL. No way to find the latter without consulting the density tables (or determining it experimentally).

Also, speaking about 'volume' of H₂SO₄ in the context of a diluted solution doesn't make much sense. Yes, you need around 5.36 mL of a pure acid to prepare the solution mentioned. No, it doesn't mean it still occupies this volume after being diluted.

[Fish200398]

Not really get 100% sir. But it must do experiment to understand this fully