calculate the molality of a solution that has 1.02kg sucrose C12H22O11 dissolved in 554g water?
mols sucrose = grams/molar mass = ?
Then m = mols/kg H2O
molality=molesSolute/kgsolvent
= (1020/molmassSucrose)/.554
in my head, on the order of m=7
To calculate the molality of a solution, you need to know the mass of the solute (in this case, sucrose) and the mass of the solvent (water).
1. Firstly, convert the given masses of the solute and solvent to grams if they are not already in grams:
- 1.02 kg sucrose = 1.02 kg * 1000 g/kg = 1020 g sucrose
- 554 g water (already in grams)
2. Next, calculate the molality using the formula:
Molality = moles of solute / mass of solvent (in kg)
3. Calculate the moles of sucrose by dividing the mass of sucrose by its molar mass:
- The molar mass of sucrose (C12H22O11) can be calculated by summing up the individual atomic masses of carbon (C), hydrogen (H), and oxygen (O).
- Carbon (C) has a molar mass of 12.01 g/mol, hydrogen (H) has a molar mass of 1.01 g/mol, and oxygen (O) has a molar mass of 16.00 g/mol.
- Therefore, the molar mass of sucrose is (12.01 * 12) + (1.01 * 22) + (16.00 * 11) = 342.34 g/mol.
Moles of sucrose = mass of sucrose / molar mass of sucrose
4. Convert the mass of water to kilograms:
Mass of water = 554 g * (1 kg / 1000 g) = 0.554 kg
5. Finally, substitute the calculated values in the molality formula:
Molality = moles of sucrose / mass of water (in kg)
Calculate the moles of sucrose using the value obtained in step 3 and substitute it into the formula:
Molality = (mass of sucrose / molar mass of sucrose) / mass of water (in kg)
Molality = (1020 g / 342.34 g/mol) / 0.554 kg
Simplifying the expression:
Molality = (2.98 mol) / (0.554 kg)
Calculating the final value:
Molality ≈ 5.37 mol/kg
Therefore, the molality of the solution is approximately 5.37 mol/kg.