Question

A student has 540.0 mL of a 0.1476 M aqueous solution of Na₂SO₃ to use in an experiment. She accidentally leaves the container uncovered and comes back the next week to find only a solid residue. The mass of the residue is 20.10 g. Determine the chemical formula of this residue.

Answer 1

ANSWETR:

Volume of Na₂SO₃ = 540.0 mL = 0.54 L

molarity of Na₂SO₃ = 0.1476 M

Molar mass of Na₂SO₃ = 126.043 g/mol

Molarity = (weight of Na₂SO₃)/(molar weight of Na₂SO₃ x volume of solution in L)

0.1476 M = (weight of Na₂SO₃)/(126.043 g/mol x 0.54 L)

weight of Na₂SO₃ = 10.05 g

Now,

Given weight of residue = 20.10 g

When we divide given weight of residue by weight of Na₂SO₃. We get

(20.10 g)/(10.05 g) = 2

That means given residue is dimer of Na₂SO₃.

Hence, the chemical formula of this residue is Na₄S₂O₆.