Which solution is more concentrated 1.0 M or 1.0 m?

Question

If WE TAKE AQUOUS SOLUTIONS .

M = Molarity
m = Molality

Answer 1

What? That's crazy, something smaller is almost always more concentrated.
What makes up a being, entity, atom, etc.. has to be squished into something smaller, or it has to have peripherals that are merely an extention, not a particle of being. So m is more concentrated than M, if it is the same.

Answer 2

Let me answer through an example.

When you dissolve a compound like glucose in water, usually the density of the solution slightly increases compared to the density of pure water. (Some compounds decrease the density, like methanol, but again for moderate concentrations the effect is very small).

For the sake of simplicity let's assume that the change in density is practically 0 and that dsolution= dwater= 1kg /L
Let's assume that the compound in question is glucose.
Glucose has a molecular weight of 180.
Thus 1 M means that you have 1 mole/ 1Lsolution.
d=1kg/L thus it is 1 mole / 1kg of solution.
But in 1 kg of solution you have glucose 1mole=180 g, thus the amount of solvent is 1-0.18= 0.82 kg.
So 1 M corresponds to 1/0.82= 1.2 m
So 1 M is more concentrated than 1 m.

Answer 3

Well molality is moles/kg and molarity is moles/L. Since density of water equals 1 g/mL, I think concentration will be the same.

Answer 4

this question doesn't make sense. Should the second one be 1.0 mM?

In that case, then 1.0M is stronger since the small m means 'milli' M or 1/1000th of a M.

Answer 5

Lancenigo di Villorba (TV), Italy

I AGREE NOT SOMEONE AMONG THE PREVIOUS ANSWERS.
I CONFIRM BELLEROPHON's one, SO HE SMILES.

!!!SURELY, "1M" solution contains more solute than "1m" one!!!

DEFINITIONS
"1 M" solution means ONE MOLE of SOLUTE in ONE LITER of SOLUTION.
"1 m" solution means ONE MOLE of SOLUT in ONE KILOGRAM of SOLVENT.

RELATIONSHIP
On the basis of the Solution's Density I can relate the two expressions.
I start from the several assumpts verified in the case of solutions :

i) ADDITIVITY OF THE MASS in the solution, that is :
SOLUTE's mass + SOLVENT's mass = SOLUTION's mass,
MW * cM * V / 1,000 + SOLVENT's mass = rho * V,
hence I obtain SOLVENT's mass = (rho - MW * cM / 1,000) * V

ii) INVARIANCE OF THE MOLE's NUMBER, that is :
cM * V = SOLUTE's moles = cm * SOLVENT's mass,
hence cM * V = cm * (rho - MW * cM / 1,000) * V

CONCLUSIONS
Finally, I rearrange the terms :

cM = cm * rho / (1 + cm * MW / 1,000)

which is an "Hyperbolic Curve" in a 'cM vs. cm's Chart'.
What is the "Asymptotic Limit" of this curve?
cM --> 1,000 * rho / MW
THUS I STATED MOLARITY's values OBEYS TO A SUPERIOR LIMIT ON THE BASIS OF ITS DEPENDANCE OF MOLALITY's values.
ON OTHER TERMS, WHATEVER YOU CHOOSE A MOLALITY's value, A DISCRETE MOLARITY's value is related to MOLALITY's one.
INDUCTION's rules LEAD TO AFFIRM 1M IS GREATER THAN 1m.

I hope this helps you.