What mass of ethanoic acid would you used and how would you prepare 0,500 dm3 of a 0,500 mol dm3 ethanoic acid solution? (M of ethanoic acid=60,0)
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What mass of ethanoic acid would you use and how would you prepare 0,500 dm3 of a 0,500 mol/ dm3 ethanoic acid solution? (molecular mass of ethanoic acid=60,0 g/mole)?
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(0,500 dm3)(0,500 mol/ dm3) = 0,250 mol ethanoic acid.
(0,250 mol)(60,0g/mol) = 15,0 grams of ethanoic acid.
Preparation:
*Weigh 15,0g of pure ethanoic acid OR calculate the volume by dividing 15,0g by its density.
*Transfer quantitavily the ethanoic acid into a 500,0 mL volumetric flask.
*Add distilled water up to the mark.
*Place the stopper firmly put you hand over it and mix by inverting 20 times.
*Transfer into a clean and labeled flask.
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Please note: I tried hard to use commas in the place of decimal points which is the custom here (USA). I gave you a more detailed answer than we usually do. Thank you for using our boards.
To calculate the mass of ethanoic acid needed to prepare a 0.500 mol/dm3 solution in a volume of 0.500 dm3, you can use the equation:
Moles = Concentration x Volume
Given:
Concentration = 0.500 mol/dm3
Volume = 0.500 dm3
Substituting these values into the equation, we can solve for the moles:
Moles = 0.500 mol/dm3 × 0.500 dm3
Moles = 0.250 mol
To find the mass of ethanoic acid, use the equation:
Mass = Moles x Molar mass
Given:
Moles = 0.250 mol
Molar mass = 60.0 g/mol
Substituting these values into the equation, we can solve for the mass:
Mass = 0.250 mol × 60.0 g/mol
Mass = 15.0 g
Therefore, you would need 15.0 grams of ethanoic acid to prepare 0.500 dm3 of a 0.500 mol/dm3 ethanoic acid solution.
To determine the mass of ethanoic acid needed to prepare a 0.500 mol dm3 ethanoic acid solution with a volume of 0.500 dm3, you need to use the formula:
n = C × V
where:
n = number of moles
C = concentration (mol dm3)
V = volume (dm3)
Given:
C = 0.500 mol dm3
V = 0.500 dm3
Substituting the values into the formula, we get:
n = 0.500 mol dm3 × 0.500 dm3 = 0.250 moles
Now, since we know the molar mass of ethanoic acid (CH3COOH) is 60.0, we can use the equation:
mass = n × molar mass
Substituting the values, we get:
mass = 0.250 moles × 60.0 g/mol = 15.0 grams
Therefore, you would need to use 15.0 grams of ethanoic acid to prepare 0.500 dm3 of a 0.500 mol dm3 ethanoic acid solution.