What is the molarity of an aqueous solution that is 6.44% glucose (C6H12O6) by mass? (Assume a density of 1.03 g/mL for the solution.)
6.44% by mass means 6.44 g glucose/100 g solution. What is the volume of the solution? That is mass = volume x density. You know d and you know mass; solve for volume in mL and convert to L.
Then find mols solute. That is 6.44g/molar mass glucose.
Finally, M = mols/L solution. You know mols and L solution.
To determine the molarity of the aqueous solution, we need to know the amount of solute (glucose) in moles and the volume of the solution in liters.
First, let's calculate the amount of glucose (C6H12O6) in grams present in the solution. We can assume we have 100 grams of the solution, as the mass percentage is given.
Mass of glucose = Mass percentage of glucose × Mass of the solution
Mass of glucose = 6.44% × 100 g
Mass of glucose = 6.44 g
Next, we need to convert the mass of glucose to moles. We can use the molar mass of glucose to make this conversion.
Molar mass of glucose (C6H12O6) = (6 × atomic mass of carbon) + (12 × atomic mass of hydrogen) + (6 × atomic mass of oxygen)
Molar mass of glucose (C6H12O6) = (6 × 12.01 g/mol) + (12 × 1.01 g/mol) + (6 × 16.00 g/mol)
Molar mass of glucose (C6H12O6) = 180.18 g/mol
Moles of glucose = Mass of glucose / Molar mass of glucose
Moles of glucose = 6.44 g / 180.18 g/mol
Moles of glucose ≈ 0.0357 mol
Now, we need to determine the volume of the solution in liters. We can use the given density of the solution to find the volume.
Density = Mass / Volume
Volume = Mass / Density
Volume = 100 g / 1.03 g/mL
Volume ≈ 97.09 mL
Now, we need to convert the volume from milliliters to liters.
Volume = 97.09 mL × (1 L / 1000 mL)
Volume ≈ 0.0971 L
Finally, we can calculate the molarity of the solution using the formula:
Molarity = Moles of solute / Volume of solution (in liters)
Molarity = 0.0357 mol / 0.0971 L
Molarity ≈ 0.368 M
Therefore, the molarity of the aqueous solution is approximately 0.368 M.