How many grams of will be needed to produce 575mL of a solution that has a concentration of Na^+ions of 1.40 M?
mass of Na3PO4 =________
I have tried & still don't get the correct answer.
2007-10-15 05:42:27 · 5 answers · asked by cvkl 1 in Science & Mathematics ➔ Chemistry
1.4M sodium means 1.4 moles of sodium/liter, but since you want only 575 ml, that is 0.575 liters, so 1.4mol/L x 0.575 L
= 0.805 moles of Na ion needed. Since each mole of Na3PO4 provides 3 moles of Na ion, we need only 1/3 of the 0.805 moles or 0.268 moles of Na3PO4. The molar mass of Na3PO4 is approximately 164 (rounded off and not using the hydrated form), so 164gm/mole x 0.268 moles = 43.95 grams
2007-10-15 06:04:59 · answer #1 · answered by Simonizer1218 7 · 3⤊ 1⤋
the mass of Na3PO4 = 3*23 (Na3) + 31 (P) + 4*16 (O4)= 164
1.4 M Na+= x moleNa/ 0.575 L solve for x... x = 0.805 moleNa+ needed for the solution..
0.805 moleNa+(1mole Na3PO4/3 moleNa+)(164gNa3PO4/1moleNa3PO4)= 44 gNa3PO4
2007-10-15 06:28:25 · answer #2 · answered by Allen C 3 · 0⤊ 1⤋
Na3po4
2016-10-06 04:54:22 · answer #3 · answered by ? 4 · 0⤊ 0⤋
conccentration= No. of moles/volume(L)
1.40 = X/0.575
so x = 1.40 * 0.575 = 0.805 mole Na3PO4
needed
now converting to (g)
mass= no.moles * m.w
= 0.805 * 146
=117.53 g of Na3PO4 needed.
2007-10-15 06:05:31 · answer #4 · answered by sami_dodeen 3 · 0⤊ 3⤋
1.4*0.575*(69+31+64) g
2007-10-15 06:00:19 · answer #5 · answered by ag_iitkgp 7 · 0⤊ 0⤋