Answer to Question #164426 in General Chemistry for bet

Question

The density of phosphoric acid (H3PO4) is 1.885 g/cm². How many moles are present in 0.245 L of phosphoric acid?

Answer 1

Given:

The density of phosphoric acid (H3PO4) is 1.885 g/cm². How many moles are present in 0.245 L of phosphoric acid?

To find: The number of moles being present in 0.245 L of phosphoric acid.

Calculation:

The density units are expressed in g/cm³ and not in g/cm².

The Density of H₃PO₄ = 1.885 g/cm³ = 1.885 g/mL                      

Since it is known that 1 g/mL = 1 g/cm³

The volume of H₃PO₄ = 0.245 L = 245 mL                                 

Since it is known that 1 L = 1000 mL

The formula for mass is given as,

mass = density X volume

The mass of H₃PO₄ is calculated as below,

=> Mass of H₃PO₄ = 1.885 X 245 = 461.825 g

The formula for molar mass is given as,

Molar mass of H₃PO₄ = Atomic mass of H X 3 + Atomic mass of P + Atomic mass of O X 4 = 1 X 3 + 31 + 16 X 4 = 98 g/mol.

Since mass = moles X molar mass

=> 461.825 = moles of H₃PO₄ X 98

=> Moles of H₃PO₄ = 4.7125 mol.         

Hence 4.7125 moles of H₃PO₄ are present in 0.245 L of H₃PO₄.