TOPICAL ASSINMENT 2 1. If 25cm' of O. IM H, SO, solution neutralised a solution containing 1.06g anhydrous sodium carbonate in
250cm. of solution, calculate the molarity and the volume of the sodium carbonate solution used. (3mks)

The molarity = 0.04 M

The volume = 6.25 ml

Given

25cm³ of 0.1 M H₂SO₄

1.06g Na₂CO₃ in  250cm³ of solution

Required

the molarity and the volume of Na₂CO₃

Solution

Molarity of  Na₂CO₃ :

Reaction

H₂SO₄ + Na₂CO₃ ⇒Na₂SO₄ + H₂O + CO₂

mol H₂SO₄ = 25 x 0.1 = 0.25 mlmol= 2.5 x 10⁻⁴ moles of  Na₂CO₃

The volume of solution used :

the electron configuration that you gave me has 16 electrons ergo it has 16 protons ergo it is the element sulfur.