You make 1.000 L
of an aqueous solution that contains 35.0 g
of glucose (C6H12O6)
.
To determine the molarity of the glucose solution, we first need to calculate the molar mass of glucose.
The molar mass of glucose (C6H12O6) can be calculated as follows:
(6 x atomic mass of C) + (12 x atomic mass of H) + (6 x atomic mass of O)
= (6 x 12.01 g/mol) + (12 x 1.01 g/mol) + (6 x 16.00 g/mol)
= 72.06 g/mol + 12.12 g/mol + 96.00 g/mol
= 180.18 g/mol
Now, we can calculate the number of moles of glucose in 35.0 g:
35.0 g / 180.18 g/mol = 0.1944 mol
Since we have 1.000 L of solution, we can calculate the molarity:
Molarity = moles of solute / liters of solution
Molarity = 0.1944 mol / 1.000 L = 0.1944 M
Therefore, the molarity of the glucose solution is 0.1944 M.