3.88 g of monoprotic acid was dissolved in water and the solution made up 250cm3. 25cm3 of this solution was titrated with .095 M NaOH solution, requiring 46.5cm3. Calculate the relative molecular mass of the acid.
2006-11-09 07:54:33 · 4 answers · asked by Joyce 4 in Science & Mathematics ➔ Chemistry
Cool stuff, I used to get confused by ugly chemistry calculations, too :D
You have 3.88 g acid in 250cm3, so in the 25 cm3 you titrated there are 0.388 g of the acid. You need to know how many moles that is, so that you can calculate molar mass.
The amount of NaOH used up is 46.5 cm3 × 0.095 mol/dm3, that is 0.0465 dm3 × 0.095 M, which equals 0.00442 mol (It's not exactly that much, but I'll stick with three significant figures).
Since NaOH reacts with monoprotic acids in a 1:1 ratio, that means that there was exactly the same amount of acid in the solution.
Now we know that 0.388 g of the acid is 0.00442 mol, so we can divide the mass by the number of moles to get molar mass. That is 0.388g/0.00442 mol = 87.8 g/mol.
Relative molecular mass is simply molar mass in grams, but without the unit, so Mr of the acid is 87.8.
2006-11-09 08:06:20 · answer #1 · answered by Anonymous · 1⤊ 0⤋
Suppose MA = monoprotic acid. The answer is g MA/mol MA. You already have 3.88g MA in the given. So you need something that will turn into mol MA.
Begin with 3.88g MA/46.5mL NaOH. Multiply by 1000mL NaOH/0.095mol NaOH. That comes from the given concentration of NaOH solution. The mLs NaOH cancel, giving mols NaOH. Next multiply by 1mol NaOH/1mol MA. That comes from the balanced equation for the reaction of a monoprotic acid with NaOH, 1:1. The mols NaOH cancel, giving gMA/mol MA.
2006-11-09 16:05:42 · answer #2 · answered by steve_geo1 7 · 0⤊ 0⤋
21.96g
2006-11-09 16:10:02 · answer #3 · answered by bige1236 4 · 0⤊ 0⤋
AAAAAHHHHH
2006-11-09 15:58:12 · answer #4 · answered by Anonymous · 0⤊ 0⤋