What is the molar concentration of sodium in a 200mL solution prepared from 1.359 grams of sodium phosphate?
To calculate the molar concentration of sodium in a solution, we need to know the number of moles of sodium and the volume of the solution in liters.
First, let's calculate the number of moles of sodium phosphate. We can use its molar mass to convert the given grams of sodium phosphate to moles.
The molar mass of sodium phosphate (Na3PO4) can be calculated by adding the atomic masses of sodium, phosphorus, and oxygen together:
Na = 22.99 g/mol (sodium)
P = 30.97 g/mol (phosphorus)
O = 16.00 g/mol (oxygen)
Molar mass of Na3PO4 = (3 * Na) + P + (4 * O)
= (3 * 22.99) + 30.97 + (4 * 16.00)
= 69.00 + 30.97 + 64.00
= 163.97 g/mol
Now, we can calculate the number of moles of sodium phosphate:
Number of moles = Mass of substance / Molar mass
= 1.359 g / 163.97 g/mol
≈ 0.00828 mol
Since sodium phosphate has a 1:1 molar ratio with sodium, the number of moles of sodium in the solution is also 0.00828 mol.
Next, we need to determine the volume of the solution in liters. The given volume is 200 mL, which needs to be converted to liters:
Volume = 200 mL × (1 L / 1000 mL)
= 0.200 L
Finally, we can calculate the molar concentration of sodium (Na) in the solution:
Molar concentration (M) = Number of moles / Volume in liters
= 0.00828 mol / 0.200 L
≈ 0.0414 M
Therefore, the molar concentration of sodium in the 200 mL solution prepared from 1.359 grams of sodium phosphate is approximately 0.0414 M.
Na3PO4, figure the moles of it from 1.359grams.
Then, you have three times that number of moles for the moles of sodium ions.