Question

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 300 mL of a solution that has a concentration of 1.10 M?

A) 40.8 grams
B) 15.8 grams
C) 16.6 grams
D) 31.6 grams

Answer 1

To make a 1.10 M solution of Na3PO4 with a volume of 300 mL, you will need 54 grams of Na3PO4.

Step-by-step explanation:

To calculate the number of grams of Na3PO4 needed, we can use the formula:

Number of grams = (Molarity * Volume * Formula mass) / 1000

Given that the volume of the solution is 300 mL and the concentration is 1.10 M, we can calculate the number of moles of Na3PO4 using the formula:

Number of moles = Concentration * Volume

Once we have the number of moles, we can multiply it by the formula mass of Na3PO4 to get the number of grams needed. The formula mass of Na3PO4 is 163.94 g/mol.

Using the given values, we can substitute them into the formulas:

Number of moles = 1.10 M * 0.300 L = 0.330 mol

Number of grams = (0.330 mol * 163.94 g/mol) / 1 mol = 54 grams

Therefore, the correct answer is A) 54 grams.