what mass of sucrose, C12H22O11, is needed to make 300.0 mL of a .50M solution?
To determine the mass of sucrose needed to make a 0.50 M solution in 300.0 mL, we can follow these steps:
Step 1: Convert the volume to liters.
300.0 mL ÷ 1000 = 0.300 L
Step 2: Use the formula c = n/V to calculate the number of moles (n) of sucrose.
0.50 M = n/0.300 L
Rearranging the formula to solve for n:
n = 0.50 M × 0.300 L
n = 0.15 moles
Step 3: Calculate the molar mass of sucrose (C12H22O11).
The molar mass of carbon (C) = 12.01 g/mol
The molar mass of hydrogen (H) = 1.01 g/mol
The molar mass of oxygen (O) = 16.00 g/mol
Molar mass of C12H22O11 = (12.01 g/mol × 12) + (1.01 g/mol × 22) + (16.00 g/mol × 11)
Molar mass of C12H22O11 = 144.12 g/mol + 22.22 g/mol + 176.00 g/mol
Molar mass of C12H22O11 = 342.34 g/mol
Step 4: Calculate the mass (m) of sucrose using the formula m = n × M.
m = 0.15 moles × 342.34 g/mol
m = 51.35 g
Therefore, 51.35 grams of sucrose (C12H22O11) is needed to make a 0.50 M solution in 300.0 mL.
To find the mass of sucrose needed to make a solution, you need to know the molar mass of sucrose and the desired molarity of the solution.
1. First, let's calculate the number of moles needed for a 0.50 M solution. Molarity (M) is defined as moles of solute per liter of solution.
Molarity = Moles of solute / Volume of solution in liters
Rearrange the equation to solve for moles of solute:
Moles of solute = Molarity x Volume of solution in liters
Moles of solute = 0.50 mol/L x 0.300 L = 0.15 moles
2. The molar mass of sucrose, C12H22O11, can be calculated using the atomic masses of carbon (C), hydrogen (H), and oxygen (O).
Molar mass of C12H22O11 = (12.01 g/mol x 12) + (1.01 g/mol x 22) + (16.00 g/mol x 11)
Molar mass of C12H22O11 = 342.34 g/mol
3. Finally, use the moles of sucrose and its molar mass to find the mass needed.
Mass of sucrose = Moles of sucrose x Molar mass of sucrose
Mass of sucrose = 0.15 mol x 342.34 g/mol ≈ 51.35 g
Therefore, you would need approximately 51.35 grams of sucrose to make a 0.50 M solution in 300.0 mL of solution.
M = moles/L
You know M and L, solve for moles.
Then moles = grams/molar mass. You know moles and molar mass, solve for grams.