Final answer:
The molarity of a 10.0% glucose solution with a density of 1.0375 g/cm³ is calculated to be 0.5759 M by first determining the mass of glucose per liter, converting it to moles, and then dividing by the volume of the solution.
Step-by-step explanation:
To calculate the molarity of a 10.0% glucose (C6H12O6) solution with a density of 1.0375 g/cm3, we will begin by determining the mass of glucose in 1 liter (1000 mL) of solution. This is done by first calculating the mass of the solution and then finding the mass of the glucose based on the percentage concentration.
First, we find the mass of 1 liter of solution:
- (1.0375 g/cm3) × (1000 cm3/L) = 1037.5 g/L
Now, since the solution is a 10% glucose solution, 10% of this mass is glucose:
- (10/100) × 1037.5 g/L = 103.75 g of glucose
To convert the mass of glucose to moles, we use the molar mass of glucose, which is approximately 180.16 g/mol:
- (103.75 g) / (180.16 g/mol) = 0.5759 mol of glucose
The molarity is the number of moles of solute per liter of solution:
- 0.5759 mol / 1 L = 0.5759 M
Therefore, the molarity of the glucose solution is 0.5759 M.