Question

What will be the normality of a solution containing 4.9 g of h3po4?

Answer 1

We are given a solution of H3PO4 having 4.9% of H3PO4 by mass .

Therefore ,

100 g of the solution contains = 4.9 g H3PO4

We know the equation for density of a substance is the mass of the substance divided by the volume of the substance

Density (ρ) = mass(m)/volume(V)

Therefore ,

V = m/ρ

Therefore , volume of H3PO4 = (4.9/1.22) mL

Volume of H3PO4 = 4.01 mL ~ 4 mL

Also ,

100 g of the solution contains = (100 - 4.9) g water

100 g of the solution contains = 95.1 g water .

We know the density of water is 1 g/mL

So , Volume of water = (95.1/1) mL

Volume of water = 95.1 mL

Therefore , the total volume of the solution is = (95.1 + 4)mL

Volume of the solution = 99.1 mL

Therefore ,

99.1 mL of the solution contains = 4.9 g H3PO4

1 mL of the solution contains = (4.9/99.1)g H3PO4

1000 mL of the solution contains = {(4.9 * 1000)/99.1} g H3PO4

1000 mL of the solution contains = 49.434 g H3PO4

Now ,

The molecular mass of phosphoric acid (H3PO4) is = {(1*3) + (31) + (16*4)}

Molecular mass = {3 + 31 + 64}

Molecular mass = 98 g

Therefore the number of moles of phosphoric acid present in 1000 mL of the solution is = (49.434/98) moles

Number of moles = 0.504 moles ~ 0.5 moles

Therefore the solution is 0.5 Molar(M) because molarity is defined as the number of moles of solute present in 1000 mL of the solution .