Answer to Question #102801 in General Chemistry for sheng

Question

Arrange the following solutions in (a) order of decreasing freezing point and in (b)
order of increasing boiling point: 0.10

Answer 1

Well the question seems to be incomplete but let's see how to deal these type of questions

Suppose we are given

Q.)Arrange the following aqueous solutions in order of decreasing freezing points (lowest to highest temperature)$0.10 m\ Na_3PO_4,\ 0.35 m\ NaCl,\ 0.20 m\ MgCl_2,\ 0.15 m\ C_6H_{12}O_6 \ and \ 0.15 m\ CH_3COOH.$

Solution

$\Delta T=i\times K_f\times m$

where i is the van't Hoff factor which is 4 for Na₃PO₄, 2 for NaCl, 3 for MgCl₂, 1 for C₆H₁₂O₆ and 1 for CH₃COOH

basically here we have to see Larger the concentration of the concentration of the particles, smaller the freezing point.

0.10m Na₃PO₄

Na₃PO₄ → 3Na⁺ + PO₄^3-

⇒ van't Hoff factor is 3 + 1 = 4

0.10 m * 4 = 0.40

0.35m NaCl

NaCl → Na+ + Cl-

⇒ van't Hoff factor = 1+1 = 2

0.35 * 2 = 0.70

0.20m MgCl₂

MgCl₂ → Mg²⁺ + 2Cl^-

 ⇒ Van't Hoff factor = 1+2 = 3

0.20 * 3 = 0.60

0.15m C₆H₁₂O₆

⇒ for non-ionic compounds in solution, like glucose (C₆H₁₂O₆) , the van't Hoff factor is 1. They do not dissociate in water.

0.15 * 1 = 0.15

0.15m CH₃COOH.

CH₃COOH ⇄ CH3COO^- + H⁺

 ⇒ Van't hoff factor ≈ 1<x<2

0.15 * 2 = 0.30 to 0.15

Larger the concentration of the concentration of the particles, smaller the freezing point.

0.15 m C₆H₁₂O₆ > 0.15 m CH₃COOH > 0.10 m Na₃PO₄ > 0.20 m MgCl₂ > 0.35 m NaCl