Calculate the molality (in mol kg-1) of C2H5OH in a water solution that is prepared by mixing 29.73 mL of C2H5OH with 524.9 mL of H2O at 20°C. The density of C2H5OH at 20°C is 0.789 g/mL.
Any help is greatly appreciated!
Any help is greatly appreciated!
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Molality is defined as (moles of solute/kg of solvent)
You must start with the density of C2H5OH, ethyl alcohol, given and multiply it by the volume, 29.73 mL to then obtain the mass of ethyl alcohol you have. Your set up should appear as follows:
0.789 g/mL = (x/29.73)
Multiply each side by 29.73 mL in order to isolate the x and obtain a mass of 23.46 g.
You can then take this mass value and use stoichiometry to obtain your moles of methyl alcohol. Work is as follows:
23.46 g * (1 mol/46.06 g C2H5OH) = 0.51 moles of ethyl alcohol
You then need to know that the density of water at 20 degrees C is 0.9982 g/mL.
Do the same work as prior to obtain the grams of water you have. Divide the mass you get, which should be 523.95 g, by 1000. This will give you kg. 0.524 kg.
Then divide the 0.51 mole of C2H5OH (Solute) by the 0.524 kg of H2O (Solvent) and you should get a molality of 0.973 mol/kg.
You must start with the density of C2H5OH, ethyl alcohol, given and multiply it by the volume, 29.73 mL to then obtain the mass of ethyl alcohol you have. Your set up should appear as follows:
0.789 g/mL = (x/29.73)
Multiply each side by 29.73 mL in order to isolate the x and obtain a mass of 23.46 g.
You can then take this mass value and use stoichiometry to obtain your moles of methyl alcohol. Work is as follows:
23.46 g * (1 mol/46.06 g C2H5OH) = 0.51 moles of ethyl alcohol
You then need to know that the density of water at 20 degrees C is 0.9982 g/mL.
Do the same work as prior to obtain the grams of water you have. Divide the mass you get, which should be 523.95 g, by 1000. This will give you kg. 0.524 kg.
Then divide the 0.51 mole of C2H5OH (Solute) by the 0.524 kg of H2O (Solvent) and you should get a molality of 0.973 mol/kg.
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molarity?