# Whose concentration is more 1 molar or 1 molal?

I was reading about concentration terms like molarity, molality, etc. and then I came across a question: which concentration is more:

• 1M(molar) of solute x dissolved in water vs.
• 1m(molal) of solute x dissolved in water

When I Google it, I did not find exact answers but did find some related information. My approach is:

• Convert 1M into molality and then compare if it is less or more than 1m. (?)
• The density of water is 1kg per L.
• When solute is added to it, density would definitely increase(?)
• And therefore 1L of that solution will weigh more than 1 kg.
• now if we calculate molality of 1M solution it would be less than 1m . (?)
• am I correct ??
• Density would always(?) Increase after adding solute
• **Hence 1m will always be more than 1M **
• density of water solutions can be smaller or bigger than dens. of pure water – Mithoron Feb 22 '15 at 12:33
• user14530, I think this is a good question so I took the liberty of editing the wording with the hope of making it clearer to our on-site experts here. Let me know if I've altered your meaning in any way. – Curt F. Feb 23 '15 at 5:52
• No it is more beautiful now . – 5cube Feb 24 '15 at 2:48
• One thing to be careful of is that molarity involves the volume of the whole solution, while molality involves the mass of just the solvent. – Yunfei Ma Feb 24 '17 at 16:49

To convert from molal to molar, one generally needs to know the density of solution as well as the formula weight of solute. Suppose there is a single solute $x$ in water, and that $x$ has a formula weight of $f_x$ g/mol. Suppose also that the solution of $x$ has a density of of $\rho$.

Let the molarity of $x$ in the solution be $C_x$ mol/L. Then:

• the mass concentration of $x$ in g/L is $C_x f_x$.
• the weight fraction of $x$ in grams of $x$ per total gram of solution is thus $w_x = \frac{C_x f_x}{\rho}$. The weight fraction $w_x$ must be a number between 0 and 1.
• The weight ratio of $x$ in grams of $x$ per gram of water is $\frac{\frac{C_x f_x}{\rho}}{1-\frac{C_x f_x}{\rho}} = \frac{w_x}{1-w_x}$.
• The molality is simply the weight ratio divided by the formula weight, $\frac{1000}{f_x}\frac{\frac{C_x f_x}{\rho}}{1-\frac{C_x f_x}{\rho}}$. The factor of 1000 is because molality is moles per kg of solvent, not moles per g.
• That last expression can be written as $\frac{1000}{f_x}\frac{\frac{C_x f_x}{\rho}}{1-\frac{C_x f_x}{\rho}} = \frac{1000}{f_x} \frac{w_x}{1-w_x}$

This last formula shows that molality values will always be higher than molarity values. Because $w_x$ has to be between zero and 1, then the molality formula will divide by a smaller number than then molarity formula. One molar water "dissolved" in water (say we are dissolving distilled water in tap water, or dissolving isotopically labeled water into regular water) has a molality of around 1.02.

My approach is:

• Convert 1M into molality and then compare if it is less or more than 1m. (?)

Good approach.

• The density of water is 1kg per L.

True, but as noted in the comments, the density of solutions in water can be either higher or lower than the density of pure water.

• When solute is added to it, density would definitely increase(?)

Not necessarily true.

• And therefore 1L of that solution will weigh more than 1 kg.

Not true, and this is the point where I think your logic mislead you. To make one liter of a 1 molar solution, less than one liter of pure water is usually required. This is because part of the liter of solution is taken up by solute. So not all of the liter needs to come from pure water. And since water has a density of 1 kg/L, if we need less than one liter of pure water, we need less than 1 kg of pure water. This then means that our one mole of solute is being dissolved by less than kg of solvent: thus the molality of a one-molar solution is greater than one.

The neat thing is that this is true regardless of the density or formula weight of the species being considered.

Molality is defined as the number of moles of solute per kilogram of solvent.i.e we have to weight 1kg of water (solvent) before adding solute. so 1 molal will be less concentrated.