How many moles of NaOH required to prepare 0.250L of 6.68M NaOH
also how many millimoles of benzoic acid (Fw = 122.1g) are contained in 1.5g of the pure acid??
(The second part has me especially stumped!)
2007-06-06 16:24:18 · 3 answers · asked by philip c 1 in Science & Mathematics ➔ Chemistry
1.67 moles of NaOH, from 6.68mol/L x .250L
and 12.29 millimoles because 1.5g/122.1g times 1000
2007-06-06 16:26:52 · answer #1 · answered by jrb229 1 · 0⤊ 0⤋
you want 1/4th of a liter of 6.68M NaOH. That will require 6.68/4 moles of NaOH (1.67 moles)
1.5g/122.1 = 0.0123 moles
How many millimoles in 1 mole? 1000, so
0.0123 x 1000 = 12.3 millimoles.
2007-06-06 23:32:21 · answer #2 · answered by skipper 7 · 1⤊ 0⤋
Moler = moles/liter
6.68moles x 0.250/1000 = 1.67
A millimole is 1/1000 of a mole, used for smaller amounts o'stuff.
[1.5g/122.1g/mol] x 1000mmol/mol = 12.28 mmol
2007-06-06 23:33:10 · answer #3 · answered by Flying Dragon 7 · 0⤊ 0⤋