Question

What is the molarity of a 4.9% $H_{3}P{O}_4$ solution by mass? (Density of $H_{3}P{O}_4$ solution = 1.22 g/mL)

Answer 1

Let the mass of the solution be 100 g.
So, 4.9 g of $H_{3}P{O}_4$ is present in 100 g of solution.
Moles of $H_{3}P{O}_4$
$\frac{\text{given mass}}{\text{molar mass}}=\frac{4.9}{98}=0.05$ moles

$Density=\frac{\text{mass of solution}}{\text{volume of solution}}$

⇒ volume of $H_{3}P{O}_4=\frac{100}{1.22}=81.97mL$
$Molarity=\frac{\text{moles of solute of}{H}_{3}P{O}_4}{\text{total volume of solution in mL}}\times 1000$
Molarity = $\frac{0.05}{81.97}\times 1000=0.61$ M