Oct 2, 2014
If 45.1 grams of a solute is added to 302 grams of water to make a 0.65M solution, what is the molar mass of a solute?

The answer is...
(45.1 grams)/(0.65(.302) moles)

The work given:
To determine the molar mass, the molarity equation will be needed.

Molarity = (moles of solute)/(Liters of solution)

Plugging in the values from the problem results in:

0.65 = (moles of solute)/(.302 L)

Note: The amount of the solution is given in terms of grams of water, this can be converted to mL by using the density of water (1.0 grams/mL) and then converting the mL into liters. Cross-multiplying gives the following:

0.65(.302) = moles of solute

The molar mass of a compound is in grams per mole. Therefore the molar mass of the unknown solute would be 45.1 grams of solute (from the question) divided by the number of moles just calculated 0.65(.302). Putting the pieces together results in a molar mass of the unknown solute of:

(45.1 grams)/0.65(.302)moles

My question is, isn't molarity equal to mols of solute/L of solution...
Shouldn't the liters of solution also account for the unknown solute? The equation set up 0.65=(moles of solute)/(.302L) does not account for this....

Any one else have a better way of explaining this?