What is the molality of a solution prepared by dissolving 81.5 g of ethylene glycol, HOCH2CH2OH, in 1.36 L of water? Assume the density of water is 1.00 g/ml.
What is the molality of a solution prepared by dissolving 75.2 g of ethylene glycol, HOCH2CH2OH, in 1.34 L of water? Assume the density of water is 1.00 g/ml.
mols = grams/molar mass
Solve for moles.
molality = moles/kg solvent
Solve for molaltiy
i did it & i got 0.96 it still says incorrect answer
To find the molality of a solution, you need to know the amount of solute (in moles) and the mass of the solvent (in kilograms).
First, let's find the moles of ethylene glycol (solute). The molar mass of ethylene glycol (C2H6O2) is 46.07 g/mol.
The number of moles of ethylene glycol can be calculated using the following formula:
moles = mass / molar mass
So, moles of ethylene glycol = 81.5 g / 46.07 g/mol = 1.77 mol
Next, we need to calculate the mass of water (solvent) in kilograms. The density of water is given as 1.00 g/mL, and the volume of water is 1.36 L.
The mass of water can be calculated using the formula:
mass = density x volume
mass of water = 1.00 g/mL x 1.36 L x 1000 mL/1 L = 1360 g
Now, we convert the mass of water to kilograms:
mass of water = 1360 g / 1000 = 1.36 kg
Finally, we can calculate the molality of the solution using the formula:
molality = moles of solute / mass of solvent (in kg)
molality = 1.77 mol / 1.36 kg = 1.30 mol/kg
Therefore, the molality of the solution prepared by dissolving 81.5 g of ethylene glycol in 1.36 L of water is 1.30 mol/kg.