An aqueous antifreeze solution is 31.0% ethylene glycol (C2H6O2) by mass.
The density of the solution is 1.039 g/cm3.
Calculate the molality of the ethylene glycol.
100-31.0=69.0g, or better yet, 0.069 kg of solvent.
31g of C2H6O2*(1 mole/62.06810 g)= moles of C2H6O2
m=molality=moles of C2H6O2/0.069 kg of solvent
31% means 31 g glycol/100 g soln.
Convert 31 g glycol to mols. mol = grams/molar mass
Convert 100 g solution to g solute + g solvent. Convert g H2O to kg, then m = mols/kg solvent.
Awesome! Thank you
Well, it looks like we've got some math to do! Let's turn on the funny gears and calculate the molality of the ethylene glycol.
To start off, we need to convert the given percentage by mass to grams. So, if we have 100 grams of this antifreeze solution, 31 grams would be ethylene glycol.
Next, we need to calculate the moles of ethylene glycol. Since the molecular formula of ethylene glycol is C2H6O2, its molar mass is 2 * 12.01 (C) + 6 * 1.01 (H) + 2 * 16.00 (O) = 62.07 g/mol.
To find the moles, we divide the mass (31 grams) by the molar mass (62.07 g/mol), giving us 0.5 moles of ethylene glycol.
Now, we need to find the mass of water in the solution. The given density of the solution is 1.039 g/cm3, but we need to convert it to grams. Considering that 1 cm3 is equivalent to 1 mL, we have 1.039 grams of solution.
Since the antifreeze solution is 31.0% ethylene glycol by mass, the remaining mass must be water. Therefore, the mass of water is 1.039 grams - 31 grams = 1.008 grams.
Finally, we can calculate the molality. Molality is defined as the moles of solute (ethylene glycol) divided by the mass of the solvent (water) in kilograms. Since we have 0.5 moles of ethylene glycol and 1.008 grams of water, we divide by 1000 to convert grams to kilograms. Therefore, the molality is 0.5 mol / 1.008 kg = 0.497 mol/kg.
So, the molality of the ethylene glycol in this antifreeze solution is approximately 0.497 mol/kg. Keep it cool, amigo!
To calculate the molality of ethylene glycol, we first need to understand what molality means. It refers to the moles of solute per kilogram of solvent. In this case, the solute is ethylene glycol (C2H6O2) and the solvent is water.
To solve this problem, we'll follow these steps:
Step 1: Convert the mass percentage of ethylene glycol to grams.
Step 2: Calculate the mass of water in the solution.
Step 3: Convert the masses of ethylene glycol and water to moles.
Step 4: Calculate the molality of ethylene glycol.
Let's go through each step:
Step 1: Convert the mass percentage of ethylene glycol to grams.
The mass percentage of ethylene glycol is given as 31.0%. This means that in a 100 g solution, 31.0 g is ethylene glycol. So, for convenience, let's assume we have 100 g of the solution.
Therefore, the mass of ethylene glycol is 31.0 g.
Step 2: Calculate the mass of water in the solution.
Since we assumed we have 100 g of the solution, the mass of water can be calculated as the difference between the total mass of the solution and the mass of ethylene glycol.
Mass of water = Total mass of solution - Mass of ethylene glycol
Mass of water = 100 g - 31.0 g
Mass of water = 69.0 g
Step 3: Convert the masses of ethylene glycol and water to moles.
To convert the masses to moles, we need the respective molar masses of ethylene glycol and water.
The molar mass of ethylene glycol (C2H6O2) is calculated as follows:
(2 * molar mass of carbon) + (6 * molar mass of hydrogen) + (2 * molar mass of oxygen)
Molar mass of C2H6O2 = (2 * 12.01 g/mol) + (6 * 1.01 g/mol) + (2 * 16.00 g/mol)
Molar mass of C2H6O2 = 62.07 g/mol
The molar mass of water (H2O) is calculated as follows:
(2 * molar mass of hydrogen) + molar mass of oxygen
Molar mass of H2O = (2 * 1.01 g/mol) + 16.00 g/mol
Molar mass of H2O = 18.02 g/mol
Now we can calculate the moles of ethylene glycol (C2H6O2) and water (H2O):
Moles of ethylene glycol = Mass of ethylene glycol / Molar mass of C2H6O2
Moles of ethylene glycol = 31.0 g / 62.07 g/mol
Moles of water = Mass of water / Molar mass of H2O
Moles of water = 69.0 g / 18.02 g/mol
Step 4: Calculate the molality of ethylene glycol.
Molality (m) is defined as moles of solute (ethylene glycol) per kilogram of solvent (water).
To find the molality, we need to convert the mass of water to kilograms:
Mass of water (kg) = Mass of water (g) / 1000
Now we can calculate the molality:
Molality (m) = Moles of ethylene glycol / Mass of water (kg)
Substituting the values we calculated earlier:
Molality (m) = (31.0 g / 62.07 g/mol) / (69.0 g / 1000 g/kg)
Simplifying the equation:
Molality (m) = 0.499 mol / kg
Therefore, the molality of the ethylene glycol solution is 0.499 mol/kg.