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.
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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 amount of ethylene glycol in moles using its molar mass:
Molar mass of ethylene glycol (C2H6O2) = (2 * Atomic mass of carbon) + (6 * Atomic mass of hydrogen) + (2 * Atomic mass of oxygen)
Now, referring to the periodic table, we can find the atomic masses:
Atomic mass of carbon (C) = 12.01 g/mol
Atomic mass of hydrogen (H) = 1.01 g/mol
Atomic mass of oxygen (O) = 16.00 g/mol
Therefore, the molar mass of ethylene glycol (HOCH2CH2OH) is:
Molar mass = (2 * 12.01) + (6 * 1.01) + (2 * 16.00) g/mol
Once you calculate this, you'll have the molar mass of ethylene glycol.
Next, we need to calculate the mass of the solvent (water) in kilograms. We are given the volume of water, which is 1.34 L. Since the density of water is 1.00 g/mL, we can use this information to convert the volume to mass:
Mass of water = density * volume
Now, we can calculate the mass of water.
Finally, we can calculate the molality using the formula:
Molality (m) = amount of solute (in moles) / mass of solvent (in kilograms)
Substitute the values that you have calculated and solve for the molality.