MCAT practice question: A well-sealed flask found in a laboratory is said to contain a 3 M solution of an unidentified compound. The solution is measured to be 2.5 L and to have a density of 2 g/mL. The molar mass of this compound is closest to:
A. 15 g/mol
B. 67 g/mol
C. 150 g/mol
D. 667 g/mol
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I have the answer, but my book does not have a very good explanation and further confused me. I understand that two important equations for this problem include density = mass/volume and molarity = mol solute/volume of solution. Where I am most confused is how these two are related enough to get molar mass of the compound.
Thank you for your time!
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EDIT: g/mol –> g/mL
In: Chemistry
In cases like these, you can use a process known as “dimensional analysis”. Basically, you use the units, which I’ll put in [square brackets], to figure out the proper mathematical relationship.
You know you have a 3 M solution, which means that you have 3 [mols / liter]. This is one area that might be the most of your confusion; molarity is defined per liter, not per arbitrary volume.
We can work backwards. You are interested in having [grams / mol] as the unit of your answer, so you have:
[something] = ? [grams / mol]
You currently only have mols in the numerator, but you can invert your given information to put mols in the denominator:
1/3 [liters / mols] * [something] = ? [grams / mol]
Now you need a way to change liters into grams. To do this, you need to have a fraction that can put liters in the denominator and grams in the numerator. Here you use the given density (after converting it to L instead of mL):
1/3 [liters / mols] * 2000 [grams / liter] = ? [grams / mol]
1/3 [~~liters~~ / mols] * 2000 [grams / ~~liter~~] = ? [grams / mol]
1/3 [~~liters~~ / mols] * 2000 [grams / ~~liter~~] = **666.7 [grams / mol]**
To help you think about what this means, since you know that you have three mols of your compound in every liter, you know that 1/3rd of a liter contains one mol (by inverting). You also know that one liter of your compound weighs 2000g. This method is over-estimating the molar mass of your compound, as it is including the contribution of the mass of the solute in the density figure, but that’s a little beyond the scope of this. The 2.5 liters was a red herring. Hope this helps; dimensional analysis is incredibly useful!
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