What is the concentration (in M) of the glucose solution at the end of the procedure?

Question

Glucose (molar mass=180.16 g/mol) is a simple, soluble sugar. Glucose solutions are used to treat patients with low blood sugar.

Suppose you prepare a glucose solution using the described procedure.

1. Step 1: Dissolve 233.0 g of glucose in enough water to make 500.0 mL of solution.
2. Step 2: Transfer 19.4 mL of the solution to a new flask and add enough water to make 250.0 mL of dilute solution.

Answer 1

First of all, when you approach a problem like this the first thing that you do is calculate the moles of the compound you're analyzing.

Luna, I know that by know you have mastered moles calculations, but to be thorough moles=mass/molar mass or n=m/M. In our case n=233.0 g/180.16(g/mol)=1.29₃₃ mol. You will see that the units for grams cancel out, and I like to keep a few extra figures in subscripts for "safety", since we only care about sig figs at the end of the problem.

Now, since we have the moles, we can calculate the concentration (M) in step 1. M=moles/volume or M=n/V. In our case M=1.29₃₃mol/0.5 L=2.58₆₆ M. Remember concentration is measured in mol/L. What is important to understand for step 2, is that when we pipet a certain volume out of a solution the number of moles in the new volume does not change! Therefore, we can use our friendly equation M1V1=M2V2. In this case M1 is the concentration we calculated earlier, V1 is the new volume 19.4 mL and V2 is given to us to be 250 mL.

Next, we rearrange the equation for M2 and we get M2=(M1V1)/V2. It is important to note that we do not need to convert the units for the volumes (V1 and V2) to L since they cancel out. Thus, M2=(2.58₆₆ mol/L * 19.4 mL)/250 mL=0.200 M. This result makes sense, because we are taking a "high" concentration (M1) and diluting it down.

Let me know if you have any questions.