50.1
The molar mass of acetic acid is 60,05 g.
The mass of one mole of a substance is its molecular mass in grams. The molecular mass of acetic acid (CH3COOH) is: 12 + (3 x 1) + 12 + 16 + 16 + 1 = 60, therefore, 1 mole of acetic acid has a mass of 60g. (The numbers used in the calculation are the mass numbers of each element)
Let "A" be the grams of acetic acid added. Then (A)/(100 + A) = 0.04 That is A=(100 +A)x(0.04) or A-0.04A = 4 (0.96A) = 4 A = 4/(0.96) = 4.166666667 grams
If it's a 4% solution by mass, you want 4.167g of acetic acid (25g/6)
The formula of acetic acid is NOT CHCOOH but CH3COOH which also is found back in the corrected molar mass:The total molecular mass. 12.01 gC/mol + (3 *1.008) gH/mol + 12.01 gC/mol + 2*16.00 gO/mol + 1.008 gH/mol = 60.05 grams per mole
60
The molar mass of acetic acid is 60,05 g.
The mass of one mole of a substance is its molecular mass in grams. The molecular mass of acetic acid (CH3COOH) is: 12 + (3 x 1) + 12 + 16 + 16 + 1 = 60, therefore, 1 mole of acetic acid has a mass of 60g. (The numbers used in the calculation are the mass numbers of each element)
Let "A" be the grams of acetic acid added. Then (A)/(100 + A) = 0.04 That is A=(100 +A)x(0.04) or A-0.04A = 4 (0.96A) = 4 A = 4/(0.96) = 4.166666667 grams
To prepare a 0.1 N glacial acetic acid solution, calculate the required mass by multiplying 0.1 moles by the molar mass of glacial acetic acid (60.05 g/mol). Weigh out the calculated mass and add it to a clean container. Dissolve the glacial acetic acid completely by stirring it with distilled water. Transfer the solution to a 1-liter volumetric flask and dilute it to the 1-liter mark with distilled water. Mix thoroughly, label, and store the solution properly, taking necessary safety precautions when handling glacial acetic acid.
If it's a 4% solution by mass, you want 4.167g of acetic acid (25g/6)
The formula of acetic acid is NOT CHCOOH but CH3COOH which also is found back in the corrected molar mass:The total molecular mass. 12.01 gC/mol + (3 *1.008) gH/mol + 12.01 gC/mol + 2*16.00 gO/mol + 1.008 gH/mol = 60.05 grams per mole
10 percent acetic acid indicates there is .1 liters of acetic acid, which is equal to 1.75 moles, or 105 grams per liter. To make this solution, add 7 liters of the 2M solution to 1 liter of water, or 7/8 solution to 1/8 water.
C2H4O2 Find moles first. Molarity = moles of solute/Liters of solution moles C2H4O2 = molarity/Liters solution moles C2H4O2 = 0.1 M/10 L = 0.01 moles C2H4O2 -----------------------------------now, molar mass time moles 0.01 moles C2H4O2 (60.052 grams/1 mole C2H4O2) = 0.60 grams acetic acid needed --------------------------------------------
Ch3cooh- mass is 60 as 12+3+12+32+1=60
No you do not. The gram-formula mass of sodium bicarbonate is 74, while the mass of acetic acid is 60. If the mass is both 2 grams, the proportion does not work. (2gNaHCo3/74gNaHCo3) = (2gCH3COOH/60gCH3COOH) is a false statement. One could also use dimensional analysis, but because there were very little conversions, I felt that proportions were easier.
That all depends on what substance fills the liter of space.A liter of air has a small amount of mass in it (very few grams).A liter of water has more mass in it (more grams).A liter of concrete has even more mass in it (lots of grams).An empty liter has no mass in it (no grams).