find percent composition 45 grams of glucose, C6H1206 is dissolved in enough water to make 0.500 liters of solution

Answer 1

To find the percent composition of glucose in the solution, we need to first calculate the molar mass of glucose.

Molar mass of glucose (C6H12O6):
6(12.01 g/mol) + 12(1.01 g/mol) + 6(16.00 g/mol) = 180.18 g/mol

Next, we need to calculate the moles of glucose in 45 grams:
Number of moles = mass / molar mass
Number of moles = 45 g / 180.18 g/mol
Number of moles = 0.2496 moles

Now, we need the concentration of the solution in moles per liter:
Concentration = moles / volume
Concentration = 0.2496 moles / 0.500 L
Concentration = 0.4992 M

Finally, we find the percent composition of glucose in the solution:
Percent composition = (moles of solute / moles of solution) * 100
Percent composition = (0.2496 moles / 0.4992 moles) * 100
Percent composition = 50%

Therefore, the percent composition of glucose in the solution is 50%.