What is the mass of glucose required to prepare 500ml of 1M solution
0.5 liter at 1M is 0.5 mol of C6H12O6
so
0.5 * mol mass of glucose
= 0.5 * ( 6*12 + 12*1 + 6*16)
To calculate the mass of glucose required to prepare a 1M solution in 500ml of solvent (water), you need to know the molar mass of glucose.
The molar mass of glucose (C6H12O6) can be calculated by adding up the atomic masses of its constituent elements:
C: 6 * atomic mass of carbon (12.01 g/mol)
H: 12 * atomic mass of hydrogen (1.01 g/mol)
O: 6 * atomic mass of oxygen (16.00 g/mol)
Molar mass of glucose = (6 * 12.01 g/mol) + (12 * 1.01 g/mol) + (6 * 16.00 g/mol)
Calculate the molar mass of glucose to get the following result:
Molar mass of glucose = 180.18 g/mol
Now, use the formula for molarity: Molarity = moles / volume (in liters)
Rearranging the formula, we have:
moles = Molarity * volume (in liters)
Since the volume of the solution is given in milliliters, we need to convert it to liters:
Volume = 500 ml = 500/1000 = 0.5 L
Now substitute the values into the formula:
moles = 1M * 0.5 L
moles = 0.5 mol
Finally, we can find the mass of glucose:
mass = moles * molar mass
mass = 0.5 mol * 180.18 g/mol
mass = 90.09 g
Therefore, you will need 90.09 grams of glucose to prepare 500ml of a 1M solution.
To find the mass of glucose required to prepare a 500 mL solution of 1M concentration, you would first need to know the molar mass of glucose (C₆H₁₂O₆).
The molar mass of glucose can be obtained by summing up the atomic masses of all the components in its chemical formula. In this case, carbon (C) has an atomic mass of 12.01 g/mol, hydrogen (H) has an atomic mass of 1.01 g/mol, and oxygen (O) has an atomic mass of 16.00 g/mol. By adding up these values for each element in glucose, we get:
(6 x C) + (12 x H) + (6 x O) = (6 x 12.01) + (12 x 1.01) + (6 x 16.00) = 72.06 + 12.12 + 96.00 = 180.18 g/mol
Next, we can calculate the mass of glucose needed using the formula:
Mass (g) = Molarity (mol/L) x Volume (L) x Molar mass (g/mol)
Given that the desired concentration is 1M and volume is 500 mL (0.5 L), we can substitute these values into the formula:
Mass (g) = 1 mol/L x 0.5 L x 180.18 g/mol
Mass (g) = 90.09 g
Therefore, to prepare a 500 mL solution with a concentration of 1M, you would need 90.09 grams of glucose.