How would you prepare 1.00L of 0.13m solution of sugar, C12H22O11? The density of the resulting solution is 1.10g/mL.
m=molality
Use mass = volume x density to calculate the mass of 1000 mL of the final solution. Then use the definition of molality = mols/kg solution.
I like to calculate the mass, then put molality into a single equation, which is,
kg x m x molar mass = grams.
You know kg, m, and molar mass. Calculate grams.
Post your work if you get stuck or don't understand what I've done.
I don't undastnd.
To prepare a 1.00L solution with a concentration of 0.13 m of sugar (C12H22O11), you will need to determine the amount of sugar needed in moles and then calculate the mass of sugar required using its molar mass. Here's how you can do it:
1. Determine the moles of sugar (C12H22O11) needed:
- Molality (m) is defined as the moles of solute per kilogram of solvent. In this case, the molarity (M) is given as 0.13 m.
- Since the density of the solution is provided (1.10 g/mL), we can calculate the mass of the solution:
Mass of solution = volume of solution × density
= 1.00 L × 1.10 g/mL
= 1100 g
- Convert the mass of the solution to kilograms:
Mass of solution in kg = 1100 g ÷ 1000 = 1.10 kg
- Multiply the molality by the mass of the solution to get the moles of sugar:
Moles of sugar = molality × mass of solution in kg
= 0.13 mol/kg × 1.10 kg
= 0.143 mol
2. Calculate the mass of sugar (C12H22O11) needed:
- Find the molar mass of sugar (C12H22O11) by summing up the atomic masses:
12 × 12.01 g/mol (carbon) + 22 × 1.01 g/mol (hydrogen) + 11 × 16.00 g/mol (oxygen)
= 144.13 g/mol + 22.22 g/mol + 176.00 g/mol
= 342.35 g/mol
- Multiply the moles of sugar by its molar mass to get the mass of sugar needed:
Mass of sugar = moles of sugar × molar mass
= 0.143 mol × 342.35 g/mol
≈ 49.0 g
Therefore, to prepare a 1.00L solution with a concentration of 0.13 m of sugar (C12H22O11), you will need approximately 49.0 grams of sugar.