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)
(.05)X(grams of total solution) = grams of acetic acid (grams of acetic acid)/ (mol. wt. of acetic acid(=60g/mol)) = mol. acetic acid (mol. acetic acid)/ (Liters of total solution) = molarity(M)
The molar mass of acetic acid is 60,05 g.
Weigh 60 grams of Glacial Acetic acid into a 1 liter volumetric flask. Make up to the 1 liter mark with distilled water. There is now a 6% (w/v) solution.
The mass of sugar is 100 g.
If it's a 4% solution by mass, you want 4.167g of acetic acid (25g/6)
The answer is 28,81 %.
I think you meant " How many moles of acetic acid in 25 grams of acetic acid? " We will use the chemist formula for acetic acid, 25 grams C2H4O2 (1 mole C2H4O2/60.052 grams) = 0.42 mole acetic acid =================
(.05)X(grams of total solution) = grams of acetic acid (grams of acetic acid)/ (mol. wt. of acetic acid(=60g/mol)) = mol. acetic acid (mol. acetic acid)/ (Liters of total solution) = molarity(M)
The answer is 31,05 g ethanol.
60
The molar mass of acetic acid is 60,05 g.
Weigh 60 grams of Glacial Acetic acid into a 1 liter volumetric flask. Make up to the 1 liter mark with distilled water. There is now a 6% (w/v) solution.
Codeine is 299.364 grams per mole; 30 grams in 2 liters is 0.0501 moles per liter.
The mass of sugar is 100 g.
6 grams is 0.6 percent of 1,000 grams.
When salt is dissolved in water it does not increase the volume of the water. 51.3 grams of salt in 1000 grams of water is equal to .0513, or 5.13 percent salt.