Calculate molarity of a 125mL aqueous solution containing 23.5g sodium sulphate (Na2SO4)
2007-08-13 06:08:06 · 3 answers · asked by philip c 1 in Science & Mathematics ➔ Chemistry
first, convert the grams of Na2SO4 to moles. This is done by dividing 23.5g by the molar weight (how many grams in a mole). The molar weight for Na2SO4 is 142 grams/mole. So after dividing 23.5g by 142g/mole you end up with .165 moles of Na2SO4. Then, we divide this by the volume (you must convert the volume to liters first. so 125 mL would be .125L) to find the molarity. So .165 moles/.125L will give you a 1.32M solution.
2007-08-13 06:44:27 · answer #1 · answered by Anonymous · 0⤊ 0⤋
To do this, you need to know the atomic weight of sodium sulphate: add together the atomic weight of sodium multiplied by two (since there are two sodium atoms per molecule), the atomic weight of sulfur, and the atomic weight of oxygen multiplied by four. This number is the molecular weight of sodium sulphate, and also happens to be the weight in grams of 1 mole of the compound.
Now divide 23.5 g by the MW you just calculated: this will give you the number of moles you have. We'll call this #mol. The molarity of a solution is the number of moles of the substance divided by the number of Liters of the solution. A 1M solution has 1 mole per 1 Liter. The molarity of this solution is your #mol/0.125 Liter (since 125 mL = 125/1000 L). Simple math now, right?
2007-08-13 13:38:15 · answer #2 · answered by John R 7 · 0⤊ 1⤋
First calculate the molar mass of Na2SO4 which is 142.04 g/mol.
Next n = m/M = 23.5g/142.04 g/mol = 0.165446 mol
Molarity is concentration so c = n/V = 0.165446/0.125L
c = 1.3236 mol/L
2007-08-13 13:37:02 · answer #3 · answered by Anonymous · 0⤊ 1⤋