If you have an aqueous solution that is 13.5 % Na3PO4 by mass, what is the molality of Na3PO4 in the solution?

1 answer:

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Considering the definition of percentage by mass and molality, the molality of Na₃PO₄ in the solution is 0.948 $\frac{moles}{kg}$ .

Percentage by mass

The percentage by mass expresses the concentration and indicates the amount of mass of solute present in 100 grams of solution.

In other words, the percentage by mass of a component of the solution is defined as the ratio of the mass of the solute to the mass of the solution, expressed as a percentage.

The percentage by mass is calculated as the mass of the solute divided by the mass of the solution, the result of which is multiplied by 100 to give a percentage. This is:

$percentage by mass=\frac{mass of solute}{mass of solution}x100$

Molality

Molality is the ratio of the number of moles of any dissolved solute to kilograms of solvent.

The Molality of a solution is determined by the expression:

$molality=\frac{number of moles of solute}{kilograms of solvent}$

This case

Considering 100 grams as a sample of the solution, then the value of the percentage of concentration given indicates that 13.5 g correspond to Na₃PO₄.

Remember that percent concentration by mass is calculated using the mass of solute and the mass of the solution, which includes both the solute and the solvent. Then:

mass solution= mass solute + mass solvent

100 g= 13.5 g + mass solvent

100 g - 13.5 g= mass solvent

86.5 g= mass solvent

Then, you know:

number of moles of solvent= $13.5 gramsx\frac{1 mole}{163.94 grams} =0.082 moles$ being 163.94 $\frac{grams}{mole}$ the molar mass of Na₃PO₄, this is the amount of mass a substance contains in one mole.
mass of solvent= 86.5 grams= 0.0865 kg (being 1000 g= 1 kg)