let me rephrase my question. A Mole of ethyl alcohol weights 46g.How many grams of ethyl alcohol are needed to produce 1 L of a 2 milimolar solution? I know the answer is 0.092g but I don't know how to derive it
massneeded= molmass*volume*molarity
I would memorize that, it saves a lot of time when making solutions.
IT comes from
molarity= (mass/molmass)/volumesolution
To derive the answer, you need to understand a few key concepts. Let's break it down step by step.
1. First, calculate the molar mass of ethyl alcohol. Ethyl alcohol, which has the chemical formula C2H6O, is composed of two carbon atoms (C), six hydrogen atoms (H), and one oxygen atom (O). To calculate the molar mass, you multiply the atomic mass of each element by the number of atoms and add them together.
C2H6O:
(2 * atomic mass of carbon) + (6 * atomic mass of hydrogen) + (1 * atomic mass of oxygen)
Using rounded atomic masses:
(2 * 12.01 g/mol) + (6 * 1.01 g/mol) + (1 * 16.00 g/mol) = 46.07 g/mol
Therefore, the molar mass of ethyl alcohol is approximately 46.07 grams per mole.
2. Convert millimolar to molar concentration. The given concentration is 2 millimolar, which means 2 millimoles of ethyl alcohol per liter of solution. To convert millimolar to molar, divide the concentration by 1000.
2 millimolar / 1000 = 0.002 moles per liter (or 0.002 mol/L)
3. Determine the number of moles of ethyl alcohol needed to make a 2 millimolar solution. The concentration of 0.002 moles per liter means that for every liter of solution, you need 0.002 moles of ethyl alcohol.
4. Finally, multiply the number of moles by the molar mass of ethyl alcohol to find the mass in grams. By using the equation: mass = moles * molar mass.
mass = 0.002 moles * 46.07 g/mol = 0.09214 g
Rounding to the appropriate significant figures, the answer is approximately 0.092 g.
Therefore, you would need approximately 0.092 grams of ethyl alcohol to produce 1 liter of a 2 millimolar solution.