A scientist wants to make a solution of tribasic sodium phosphate,Na3PO4 , for a laboratory experiment. How many grams of Na3PO4 will be needed to produce 550 ml of a solution that has a concentration of Na+ ions of 1.40M ?
ohh that's def the nth root of go to hell and chemistry sucks
1.40M Na^+ we convert to Na3PO4 which means we need 1.40M x (1 mole Na3PO4/3 mol Na^+) = 1.40 x 1/3 = 0.467 M Na3PO4.
How many moles Na3PO4 do you need? That is M x L = 0.467 x 0.550 = ?
moles Na3PO4 = grams Na3PO4/molar mass Na3PO4. You know moles and molar mass, solve for grams.
2+SF++MOL+1MOL+3,MD
To calculate the number of grams of Na3PO4 needed, we need to use the formula:
Moles = Volume (in liters) × Concentration (in mol/L)
First, let's convert the given volume from milliliters to liters:
550 ml = 550/1000 = 0.55 L
Now we have the volume in liters. Next, we use the given concentration to calculate the moles of Na+ ions:
Moles = 0.55 L × 1.40 mol/L = 0.77 mol
Since Na3PO4 has three Na+ ions, we need to divide the calculated moles of Na+ ions by three to get the moles of Na3PO4:
Moles of Na3PO4 = 0.77 mol / 3 = 0.2567 mol
Finally, to convert moles to grams, we use the molar mass of Na3PO4. The molar mass of Na3PO4 is 163.94 g/mol.
Grams of Na3PO4 = 0.2567 mol × 163.94 g/mol ≈ 42.09 g
Therefore, approximately 42.09 grams of Na3PO4 will be needed to produce a 550 ml solution with a concentration of Na+ ions of 1.40M.