The less soluble salt : PbCl₂
<h3>Further explanation</h3>
Given
0.1 M NaCl
Required
The less soluble salt
Solution
If we see from the answer option, the salt that is more difficult to dissolve in NaCl is PbCl₂ because it has the same ion (Cl)
When PbCl₂ is dissolved in water, ionization will occur
PbCl₂ ⇒ Pb²⁺+ 2Cl⁻
So, when dissolved in NaCl, NaCl itself will be ionized
NaCl ⇒ Na⁺ + Cl⁻
Based on the principle of equilibrium, the addition of an ion (one of the ions is enlarged), the reaction will shift towards the ion that was not added. In addition to this Cl ion, the reaction will shift to the left so that the solubility of PbCl₂ will decrease (the reaction to the right decreases)
The balanced chemical reaction is written as:
Sb2S3 + 6HCl = 6SbCl<span>3 + 3H2S
We are given the amount of </span><span>antimony(III) sulfide to be used in the reaction. This is amount will be used for the calculations. We do as follows:
2.85 g Sb2S3 ( 1 mol / </span><span>339.715 g ) ( 6 mol SbCl3 / 1 mol Sb2S3 ) (</span> 228.13 g / mol ) = 11.48 g SbCl3
Answer:
ΔSv = 0.1075 KJ/mol.K
Explanation:
Binary solution:
∴ a: solvent
∴ b: solute
in equilibrium:
- μ*(g) = μ(l) = μ* +RTLnXa....chemical potential (μ)
⇒ Ln (1 - Xb) = ΔG/RT
∴ ΔG = ΔHv - TΔSv
⇒ Ln(1 -Xb) = ΔHv/RT - ΔSv/R
∴ Xb → 0:
⇒ Ln(1) = ΔHv/RT - ΔSv/R
∴ T = T*b....normal boiling point
⇒ 0 = ΔHv/RT*b - ΔSv/R
⇒ ΔSv = (R)(ΔHv/RT*b)
⇒ ΔSv = ΔHv/T*b
∴ T*b = 80°C ≅ 353 K
⇒ ΔSv = (38 KJ/mol)/(353 K)
⇒ ΔSv = 0.1075 KJ/mol.K