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