salt water solution

A chemist has m ounces of salt-water solution that is m percent salt. How much salt must be added for the solution to be 2m percent salt?

[initially:]
m ounces = m%
=> 1 ounce = 1%
=> 100 ounces = 100%
So, initially, the solution is 100 ounces

[adding in x ounces:]
m+x ounces of salt
100+x ounces of solution
=> solution = (m+x)/(100+x) * 100 = 2m
=> 100m + 100x = 200m + 2mx (multiplying through)
=> 100x - 2mx = 100m
=> x = 100m/(100-2m) = 50m/(50-m)

[solution:]
The chemist needs to add 50m/(50-m) ounces :D:wave::D

Hmm, not so because she could just as easily have a 32 ounce solution with 32% salt concentration. Whatever the value of m, she'd need to double the amount of salt in order to double it's concentration, so she'd need to add another m ounces of salt.

Oh yeh and we're all ever so sorry you didn't win the world cup... hehe...

World Cup? The only REAL world cup was in Sth. Korea and Japan, and Brazil won, Australia not even attending :p

Anyway...
if she (or he) has 32% salt concentration, that means they have 32 ounces of salt, and 68 ounces of water.
after they add 32 ounces, they will have 64 ounces of salt, 68 ounces of water. So, they will have 64/132 = 48.48% salt. This is not close to 64%...

However, if they add 32*50/(50-32) = 88.8 ounces, they would then have (120.8/188.8) = 63.9%. this is only different because i rounded 88+8/9 down to 88.8

I'm struggling to see where your 132 is coming from:

This will probably make more sense if I use grams rather than ounces (I'm sure you know one is a scalar multiple of the other so it makes no difference):

If you dissolve 32g of salt into 100g of water, you'll have a 32% solution. So if you dissolve 64g, you'll have a 64% solution. Or at least that's where I'm coming from.

The problem is that you can't really describe concentrations in terms of percentages because it's not universally defined, i.e. I say it's 32g in 100g water and you say it's 32g in 68g water. You really have to use moles and decimeters cubed:

Amount of salt (NaCl) = 32g/58.5gmol^-1 = 0.547mol

So if you dissolved that in 100cm² of water you'd get a concentration of:

0.547mol/0.1dm³ = 5.47moldm^-3

Now if you want twice that, that's 10.94moldm^-3:

Amount of salt = 10.94moldm^-3 * 0.1dm³ = 1.094 mol

Mass of salt = 1.094mol * 58.5gmol^-1 = 64g

Lo and behold that's twice as much.

And on the rugby, over here football/soccer is a ****'s sport because a lot of the fans are violent and/or oppressive. Yesterday was a very good day because rugby overtook football as the national sport, and rugby fans here are well mannered. It's the situation here over the last few years that it's unsafe to take your children to see a football match whereas they can pretty much safely take themselves to a rugby match.

If you have 32 ounces in a solution, and add another 32 ounces, then the only way this results in a doubling of percentage is if the final solution is as large as the initial solution. This means that if you add 32 ounces of salt, u have to remove some water.

If i have a drink with an ice-cube in it, and want to double the concentration, i can't just add another cube. This doubles the ice:water ratio, but not the ice:solution ration, which is what concentration is...

Interesting about your 'concentration' definition. Does this mean that 100g of salt in 100g of water is 100% solution? Or even that 150g of salt in 100g of water is 150%...hehe


Oh, and in AUS, soccer is not very popular, the crowds aren't really big enough to be violent. AFL is the national sport (i can hear hisses from NSW ppl). All Aussie soccer players are in Europe

So far as my definition of concentration goes, the concentration of an aqueous solution depends on the number of moles of the solute and the volume of the solvent. Obviously this means you can in theory have a solution with infinite concentration but in practice you can't because you won't find anything which is infinitely soluble in water.

It's just a different definition...

Suppose you want to add 100g of NaCl into 150g of water. Well what if I add 100g of PbO₄? The lead (II) oxide molecule is far larger than the sodium chloride "molecule" so your 100g of lead (II) oxide contains far less molecules (say, 300) in than your 100g of sodium chloride (say, 3000). How is it then that both concentrations don't turn out to be 100/150 = 67%, when there would be less PbO₄ molecules per molecule of water than there would be NaCl "molecules"? Your definition of concentration ignores the fact that different compounds have different molar masses, it assumes that every molecule is the same size so 100g always contains x molecules and when dissolved into 100g of water always produces a 100% solution. Quite simply, it doesn't.

Sodium chloride isn't all that soluble compared to something like sulphuric acid, so consider this situation:

You try and add 150g of NaCl to 100g of water, and some of it doesn't dissolve. Your definition of concentration jumps up out of it's seat and says "aha! you can't have a concentration of more than 100% because that's not allowed", so some doesn't dissolve. But when you try and add 150g of H₂SO₄ to 100g of water, you'll find that it does actually dissolve, producing a concentration of 15.3moldm^-3 (and that's bloody strong mind but still eminantly producable).

I'm not great at putting stuff into words but I hope that makes sense.

here is a chemists definition of concentration:
conc = moles / volumes (in dm³)

and moles = grams/Mr

In several fields, salt in water is measured in parts per thousand(ppt). Since percent simply means parts per hundred (per cent), I would say that a 32% solution is 32 parts salt per 100 parts water. However, I understand the idea that it can be seen as 32 parts salt per 100 parts of total stuff.

Interesting question, the wording has left things confused. How well do we communicate? What if it was a salt/pepper mix that was 32% salt. Then there would be no question. However, since it is a salt water solution, it all depends on the terminology.