Question

If you have an aqueous solution that is 37.0 % Na3PO4 by mass, what is the molality of Na3PO4 in the solution? Enter your answer in units of molality to three significant figures.

Answer 1

Here, we are given with the solution containing 37% of solute by mass. We need to find the molality of the solution.

Let's assume the weight of solution is 1000g i.e. 1kg. It is given that 37% of the total weight of solution is coming from the solute.

So total weight of the solute will be

w_solute = 37% of total weight of solution

so the weight of solvent will be

w_solvent = w_solution - w_solute

Molality of the solution is given by

Again no. of moles is given by

So, molality will be

Now, molar mass of Na₃PO₄ is 164g/mol.

Putting all the values in the equation of molality