Calculate the molarity of Na + ions in a 100 ml solution containing 0.46 g Na3PO4

0.084 M

Explanation:

molar mass of Na3PO4 = 3*23 + 31 + 4*16 = 164 g/mol

moles of Na3PO4 = mass / molar mass = 0.46/164 = 0.0028 mol

There are 3 moles of Na + in 1 mol of Na3PO4, so there are 3*0.0028 = 0.0084 moles of Na + in the solution

Molarity is defined as:

M = moles of solute/litre of solution

100 ml are equivalent to 0.1 L

Therefore:

M of Na + = 0.0084/0.1 = 0.084 M