# Molarity of two solutions

Question: Two solutions of ethanol marked as X and Y are labeled as 25% ethanol by mass and 25% ethanol by volume respectively. If the density of solution Y is 0.789 g/mL and that of solution X is 0.968 g/mL, identify the solution with higher molarity.

Solution :

Let total volume = $$\pu{100 mL}$$

Volume of ethanol = $$\pu{25 mL}$$

Mass of ethanol = density×volume = $$0.789×25 = \pu{19.725 g}$$

Molarity of solution Y = Mass of ethanol/Molecular mass of ethanol×1000/Volume of solution (mL)

Molarity = $$\frac{19.725}{46}\times\frac{1000}{100}=\pu{4.28 M}$$

Hence, molarity of solution Y = $$\pu{4.28 M}$$

For solution X: Let total mass of solution = $$\pu{100 g}$$

Mass of ethanol = $$\pu{25 g}$$

Volume of solution=Mass/density = $$100/0.968=\pu{103.30 mL}$$ Molarity = Mass of ethanol/Molecular mass of ethanol×1000/volume of solution (mL)

Molarity of solution X=$$\frac{25}{46}\times\frac{1000}{103.30}=\pu{5.26 M}$$

So, the molarity of solution X is more than that of Y. In the calculation of molarity of Y, density of ethanol is taken as $$\pu{0.789 g/mL}$$. But according to me, that is the density of the solution and not of ethanol. Please clarify my doubt.

Question: Two solutions of ethanol marked as X and Y are labeled as 25% ethanol by mass and 25% ethanol by volume respectively. If the density of solution Y is 0.789 g/mL and that of solution X is 0.968 g/mL, identify the solution with higher molarity.

The question is a mish-mash of data. It is a bad problem. I hate stupid problems!!

Mixture X is supposed to be 25% ethanol by mass and has a density of 0.968 g/ml.

Mixture Y is supposed to be 25% ethanol by volume and has a density of 0.789 g/ml.

According to tabular data a solution of Z% by mass should be slightly less dense than a solution of Z% by volume.

The solution with a density of 0.789 g/ml is either pure ethanol, or nearly so. (The temperature of the density data wasn't specified in the problem.) Thus mixture Y has the higher molarity of ethanol.

Now forgetting any real chemistry, let's solve the problem as given. The molecular weight of ethanol is 46.07 grams/mole.

Mixture X

25% ethanol by mass and has a density of 0.968 g/ml.

Thus 1000 ml of solution X has a mass of 968 grams. 75% of the mass is water and 25% of the mass is ethanol.

Thus the molarity is:

$$\mathrm{M}_\mathrm{X} = \dfrac{968*0.25}{46.07} = 5.25\ \mathrm{molar}$$

Mixture Y

25% ethanol by volume of solution Y has a density of 0.789 g/ml.

Thus 1000 ml has a mass of 789 grams. 75% of the volume is water and 25% of the volume is ethanol.

Now the problem can't be solved without extra data and assumptions. Perhaps the simplest is to assume that:

• Pure water has a density of 1.000 grams/ml

The density of water isn't too bad of an assumption, but assuming that the volumes are additive is a very poor assumption.

Both the concentrations of Solution X and Solution Y are correct. On the other hand, the value of the density that you used on the calculation of the mass of ethanol is also correct.

The direct relationship between the volume of ethanol on the solution and the density of the solution demonstrate the mass of the ethanol on the solution.

For example:

Mass of ethanol = (density of the solution X)(volume of ethanol) Mass of ethanol = ($$0.968 g/mL$$)($$25 mL$$) Mass of ethanol = $$24.20 grams$$

However, if you use the density of ethanol, it will result on the mass of the ethanol as ethanol itself and not as ethanol on the solution stated on the problem.

For example:

Mass of ethanol = (density of ethanol)(volume of ethanol) Mass of ethanol = ($$0.789 g/mL$$)($$25 mL$$) Mass of ethanol = $$19.73 grams$$

I use the Solution X to provide a clear comparison between the mass of ethanol using the density of the solution and the mass of ethanol using density of ethanol itself.

I think, you get confused with the calculation of the mass of ethanol using the density of Solution Y, because both of the density of ethanol and Solution Y are the same. The problem itself is not correct, it states that Solution Y is only $$25%$$ ethanol yet its density corresponds to the density of an ethanol.

The question is bad enough to put on-hold, but I'd give my take since OP tried hard to solve it.

It is given that mixture $$X$$ is 25% ethanol by mass ($$w/w$$) and has a density of $$\pu{0.968 g/mL}$$.

It is also given that mixture $$Y$$ is 25% ethanol by volume ($$v/v$$) and has a density of $$\pu{0.789 g/mL}$$.

Let's take mixture $$X$$ first:

Since it it has a density of $$\pu{0.968 g/mL}$$, $$\pu{1000 mL}$$ of this solution weighs $$\pu{968 g}$$. Yet, only 25% of this mass is ethanol.

Therefore, mass of ethanol in $$\pu{1000 mL}$$ of solution $$X = \pu{968 g} \times 0.25 = \pu{242 g}$$. Or, $$\pu{242 g} \times \frac{\pu{1 mol}}{\pu{46.1 g}} = \pu{5.249 mol}$$, since molar mass of ethanol is $$\pu{46.1 g/mol}$$.

Thus, molarity of solution $$X$$ is $$\pu{5.25 mol/L}$$.

Now, we can look at mixture $$Y$$:

Since it has a density of $$\pu{0.789 g/mL}$$ (given), we can calculate that weight of $$\pu{1000 mL}$$ of solution $$Y$$ is $$\pu{789 g}$$. However we cannot go further without knowing additional date from here as MaxW instate elsewhere. As MaxW's conclusion, we have no choice, but assume that (1) density of water is $$\pu{1.00 g/mL}$$ at operating temperature; and (2) water and ethanol are additives (meaning no contraction or expansion when mixed together).

If we can assume these two facts, then we can conclude that 25% of this $$\pu{1000 mL}$$ of solution $$Y$$ is ethanol ($$\pu{250 mL}$$) and other 75% ($$\pu{750 mL}$$) is water.

Therefore, mass of water in $$\pu{1000 mL}$$ of solution $$Y = \pu{750 mL} \times \frac{\pu{1.00 g}}{\pu{1.0 mL}} = \pu{750 g}$$. That means, mass of ethanol in $$\pu{1000 mL}$$ of solution $$Y = \pu{789 g} - \pu{750 g} = \pu{39 g}$$. Or, $$\pu{39 g} \times \frac{\pu{1 mol}}{\pu{46.1 g}} = \pu{0.846 mol}$$.

Thus, molarity of solution $$Y$$ is $$\pu{0.846 mol/L}$$.

Note: This calculation is similar to that of MaxW. Yet, just for pure curiosity, I put this answer since none of the answers have given the molarity of solution $$Y$$.