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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)​

Sagot :

ardni313

The molarity = 0.04 M

The volume = 6.25 ml

Further explanation

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 :

[tex]\tt \dfrac{1.06/106~mol}{0.25~L}=0.04~M[/tex]

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 :

[tex]\tt V=\dfrac{n}{M}=\dfrac{2.5.10^{-4}}{0.04}=6.25.10^{-3}~L=6.25~ml[/tex]

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