A student placed 13.0 g of glucose (C6H12O6) in a volumetric flask, added enough water to dissolve the glucose by swirling, then carefully added additional water until the 100. mL mark on the neck of the flask was reached. The flask was then shaken until the solution was uniform. A 40.0 mL sample of this glucose solution was diluted to 0.500 L. How many grams of glucose are in 100. mL of the final solution?

Express your answer to three significant figures and include the appropriate units.

Molarity (M) is defined as the number of moles of solute divided by the solution volume expressed in liters:

$\mathrm{Molarity}=\frac{\mathrm{moles}\mathrm{of}\mathrm{solute}}{\mathrm{v}\mathrm{of}\mathrm{solution}}$

For example, 1 M HCl contains 1 mol of HCl dissolved in 1 L of solution. When a concentrated solution is diluted, the number of moles of solute stays constant; only the volume of the solution is changed. A dilution indicates an increase in solution volume and, therefore, the concentration of the solution must decrease. If you add more water to the HCl solution considered above, so that now the volume is 2 L , the number of moles remains the same but the volume is doubled. Hence the molarity of the solution is now 1 mol in a 2 L solution, that is, (1/ 2) M or 0.5 M.

The number of moles of solute before and after dilution can be calculated by multiplying molarity times volume. We can set up the following equations:

moles of solute = molarity × volume

moles of solute = Mi × Vi = Mf × Vf

where M_{i} is the initial molarity (of the concentrated solution), V_{i }is the initial volume, M_{f} is the final molarity (of the diluted solution), and V_{f} is the final volume. In the HCl solution example the initial molarity is 1 M, the initial volume is 1 L, the final volume is 2 L, and the molarity is 0.5 M. Thus the number of moles present in these solutions is

MiVi = MfVf = 1 M × 1 L = 0.5 M × 2 L = 1 mol

Frequently Asked Questions

What scientific concept do you need to know in order to solve this problem?

Our tutors have indicated that to solve this problem you will need to apply the Dilution concept. If you need more Dilution practice, you can also practice Dilution practice problems.