How many grams of glucose are needed to make 500ml of a 3.0M solution of glucose?
mass=Molarity*volume*molmassGlucose
To determine how many grams of glucose are needed to make a 500 ml solution with a concentration of 3.0 M, you need to calculate the moles of glucose required and then convert it to grams using the molar mass.
Step 1: Convert the volume of the solution from milliliters to liters:
500 ml = 500/1000 = 0.5 L
Step 2: Use the formula for molarity to calculate the moles of glucose required:
Molarity (M) = moles of solute (glucose) / volume of solution (L)
Rearrange the formula to solve for moles of solute:
moles of solute (glucose) = Molarity (M) x volume of solution (L)
moles of solute (glucose) = 3.0 M x 0.5 L
moles of solute (glucose) = 1.5 mol
Step 3: Determine the molar mass of glucose:
The molar mass of glucose (C6H12O6) is calculated by summing the atomic masses of its elements:
(6 x atomic mass of carbon) + (12 x atomic mass of hydrogen) + (6 x atomic mass of oxygen)
= (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
Step 4: Convert moles of glucose to grams:
grams of glucose = moles of glucose x molar mass of glucose
grams of glucose = 1.5 mol x 180.18 g/mol
grams of glucose = 270.27 g
Therefore, you would need 270.27 grams of glucose to make a 500 ml solution with a concentration of 3.0 M.
To determine the number of grams of glucose needed to make a 500ml solution with a concentration of 3.0M, you need to use the formula:
Mass (grams) = Concentration (M) × Volume (liters) × Molar mass (grams/mol)
First, convert the volume from milliliters (ml) to liters (L):
Volume = 500ml ÷ 1000 = 0.5L
Next, determine the molar mass of glucose. The molecular formula for glucose (C6H12O6) consists of 6 carbon atoms (C), 12 hydrogen atoms (H), and 6 oxygen atoms (O). The atomic masses are approximately: C = 12.01 g/mol, H = 1.01 g/mol, and O = 16.00 g/mol.
Molar mass of glucose = (6 × atomic mass of C) + (12 × atomic mass of H) + (6 × atomic mass of O)
= (6 × 12.01) + (12 × 1.01) + (6 × 16.00)
≈ 180.18 g/mol
Now, substitute the known values into the formula:
Mass (grams) = 3.0M × 0.5L × 180.18 g/mol
= 270.27 g
Therefore, approximately 270.27 grams of glucose are needed to make a 500ml (0.5L) solution with a concentration of 3.0M.