calculate the molality of a solution that has 1.02kg sucrose (C12H22O11), dissolved in 554g water.
1,020 g sucrose/molar mass sucrose = moles sucrose.
molality = moles/0.554 kg solvent = ??
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To calculate the molality of a solution, you need to determine the moles of solute (sucrose) and the mass of the solvent (water). Here's how you can get the answer:
1. Calculate the moles of sucrose (C12H22O11):
- The molar mass of sucrose (C12H22O11) can be calculated by summing the atomic masses of its constituent elements. The atomic masses of carbon (C), hydrogen (H), and oxygen (O) are 12.01 g/mol, 1.01 g/mol, and 16.00 g/mol, respectively.
- The molar mass of sucrose is:
12 carbon atoms x 12.01 g/mol = 144.12 g/mol
22 hydrogen atoms x 1.01 g/mol = 22.22 g/mol
11 oxygen atoms x 16.00 g/mol = 176.00 g/mol
Total molar mass of sucrose = 342.34 g/mol
- Now we can calculate the moles of sucrose:
Moles of sucrose = Mass of sucrose / Molar mass of sucrose
Moles of sucrose = 1.02 kg x (1000 g/1 kg) / 342.34 g/mol
2. Calculate the moles of water:
- The molar mass of water (H2O) is 18.02 g/mol.
- Moles of water = Mass of water / Molar mass of water
- Moles of water = 554 g / 18.02 g/mol
3. Calculate the molality:
- Molality (m) is defined as moles of solute per kilogram of solvent. We have already calculated the moles of solute (sucrose) and the mass of solvent (water).
- Molality = Moles of solute / Mass of solvent (in kg)
- Molality = Moles of sucrose / (Mass of water / 1000)
- Plug in the values calculated in steps 1 and 2 to find the molality.
Now, you can substitute the calculated values into the formula to find the molality of the solution.