A solution is prepared by dissolving 10.00 g of glucose, C6H12O6, in 500 g of water. The
final volume of the solution is 415 mL
i)Determine the moles of glucose, C6H12O6
ii)Calculate the molality of the solution
molar mass C6H12O6 = 180
mols glucose = grams/molar mass = 10.00/180 = ?
molality = mols glucose/kg solvent = mols glucose/0.500 = ?
Step 1: Calculate the molar mass of glucose, C6H12O6
The atomic masses of carbon (C), hydrogen (H), and oxygen (O) are approximately 12.01 g/mol, 1.01 g/mol, and 16.00 g/mol, respectively.
Molar mass of C6H12O6 = (6 × molar mass of C) + (12 × molar mass of H) + (6 × molar mass of O)
Molar mass of C6H12O6 = (6 × 12.01 g/mol) + (12 × 1.01 g/mol) + (6 × 16.00 g/mol)
Molar mass of C6H12O6 = 72.06 g/mol + 12.12 g/mol + 96.00 g/mol
Molar mass of C6H12O6 = 180.18 g/mol
Step 2: Determine the moles of glucose, C6H12O6
Given mass of glucose = 10.00 g
Number of moles of C6H12O6 = mass of C6H12O6 / molar mass of C6H12O6
Number of moles of C6H12O6 = 10.00 g / 180.18 g/mol
Therefore, the moles of glucose, C6H12O6, in the solution is given by:
Number of moles of C6H12O6 = 0.055 mol
Step 3: Calculate the molality of the solution
Molality is defined as the number of moles of solute (glucose) per kilogram of solvent (water).
Given mass of water = 500 g
Volume of solution = 415 mL = 415 g (since 1 mL of water has a mass of 1 g)
Molality (m) = moles of solute / kilograms of solvent
Molality (m) = 0.055 mol / 0.415 kg
Therefore, the molality of the solution is given by:
Molality (m) = 0.132 mol/kg
To determine the moles of glucose, C6H12O6, in the solution, you need to use the given mass of glucose and its molar mass.
i) The molar mass of glucose, C6H12O6, can be calculated by adding up the atomic masses of its constituent elements. The atomic masses are:
- Carbon (C): 12.01 g/mol
- Hydrogen (H): 1.01 g/mol
- Oxygen (O): 16.00 g/mol
To calculate the molar mass of glucose, multiply the atomic masses by the number of atoms present in the formula and sum them up:
Molar mass of glucose (C6H12O6) = (6 * Carbon atomic mass) + (12 * Hydrogen atomic mass) + (6 * Oxygen atomic mass)
Molar mass of glucose = (6 * 12.01 g/mol) + (12 * 1.01 g/mol) + (6 * 16.00 g/mol)
Molar mass of glucose = 72.06 g/mol + 12.12 g/mol + 96.00 g/mol
Molar mass of glucose = 180.18 g/mol
Now, you can calculate the moles of glucose by dividing the given mass of glucose by its molar mass:
Moles of glucose = Mass of glucose / Molar mass of glucose
= 10.00 g / 180.18 g/mol
ii) To calculate the molality of the solution, you need to use the moles of solute (glucose) and the mass of the solvent (water). Molality is defined as moles of solute per kilogram of solvent.
Molality (m) = Moles of solute / Mass of solvent (in kg)
First, you need to convert the mass of water into kilograms:
Mass of water = 500 g = 500 g / 1000 g/kg = 0.500 kg
Now, you can calculate the molality:
Molality (m) = Moles of glucose / Mass of water
= Moles of glucose / 0.500 kg
Please note that the final volume of the solution provided in the question is irrelevant for calculating the molality, as it only affects the density of the solution, not the concentration of the solute.