A scientist wants to make a solution of tribasic sodium phosphate, \rm Na_3PO_4, for a laboratory experiment. How many grams of \rm Na_3PO_4 will be needed to produce 575 mL of a solution that has a concentration of \rm Na^+ ions of 1.00 \it M?
I really would like to know what \rm and
\it mean.
You want 1.0 M (I assume the \it is extraneous) Na^+ which means you want 1/3 molar Na3PO4.
M = moles/L. You know M and L, solve for moles.
Then moles = grams/molar mass. Solve for grams.
Check my thinking.
It means she was too lazy to do her homework or proofread what she is copying from her homework's website.
To determine the number of grams of Na₃PO₄ needed, we need to use the formula:
mass = moles × molar mass
Here's how you can calculate it step by step:
1. Find the moles of Na₃PO₄ needed:
The concentration of Na⁺ ions is given as 1.00 M. This means that there is 1 mole of Na⁺ ions in 1 liter (1000 mL) of the solution.
Since we have 575 mL of the solution, we can convert it to liters by dividing by 1000:
575 mL ÷ 1000 = 0.575 L
Therefore, we need 0.575 moles of Na⁺ ions.
Since Na₃PO₄ has a 1:3 ratio of Na⁺ ions to Na₃PO₄ molecules, we can multiply the number of moles of Na⁺ ions by 1/3 to find the moles of Na₃PO₄ needed:
0.575 moles × (1/3) = 0.1917 moles of Na₃PO₄
2. Find the molar mass of Na₃PO₄:
The molar mass of Na is 22.99 g/mol, and the molar mass of P is 30.97 g/mol, and the molar mass of O is 16.00 g/mol.
The molar mass of Na₃PO₄ is therefore:
(3 × 22.99) + 30.97 + (4 × 16.00) = 163.97 g/mol
3. Calculate the mass of Na₃PO₄ needed:
mass = moles × molar mass
mass = 0.1917 moles × 163.97 g/mol ≈ 31.38 grams
Therefore, approximately 31.38 grams of Na₃PO₄ will be needed to produce 575 mL of the solution with a concentration of Na⁺ ions of 1.00 M.