Question
You make 1.000 L
of an aqueous solution that contains 35.0 g
of glucose (C6H12O6)
.
of an aqueous solution that contains 35.0 g
of glucose (C6H12O6)
.
Answers
GPT 3.5
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.
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.