C ........................
Answer:
Since one mole Na₂S ionised to give 2moles Na⁺ ion, hence concentration of sodium ion in the solution is (2 × 0.1748)M = 0.3496 M
≈0.350 M
Explanation:
From the question it is clear that,
Initial volume of sodium sulphide solution is (v₁) = 50mL
Initial concentration of sodium sulphide solution is (s₁) =0.874 M
Final volume of sodium sulphide solution is (v₂) = 250mL
Let, the final concentration of sodium sulphide solution is s₂, then according to acidimetry-alkalimetry,
v₁ × s₁ = v₂ × s₂
Or, s₂ = v₁ × s₁/v₂
= 50 × 0.874 / 250
= 0.1748 M
Therefore, concentration of 250mL sodium sulphide solution is 0.1748 M
Since one mole Na₂S ionised to give 2moles Na⁺ ion, hence concentration of sodium ion in the solution is (2 × 0.1748)M = 0.3496 M
≈0.350 M
Answer:
1) Separate the aqueous layer from the organic layer using the separation funnel.
2) Treat the aqueous layer to obtain compound A.
3) Distilated the organic layer to obtain compound B.
Explanation:
When <u>NaOH is added</u> to the mixture the acid groups will react to produce a salt and increases the polarity of the compounds due to the net charges generated. (Figure 1).
Therefore, the salt produced by compound A will move to the <u>aqueosus layer</u>. Compound B dont react due to the lack of <u>acid groups</u>. So, this molecule will stay in the <u>organic layer</u>.
When the aqueous layer is separated from the organic layer using the separation funnel we will have a <u>separation</u>. The compound will remain in the aqueous layer and compound b will remain in the organic layer.
Then we aqueous layer can be <u>treated with HCl</u> in order to obtain the initial A molecule, in other words: Undo the ionic form of compound A.
The organic layer can be removed by <u>distillation</u> in order to obtain the pure form of compound B.
