Answer:
Option d.
1 mole AlCl3in 500 g water
Explanation:
ΔT = Kf . m . i
Freezing T° of solution = - (Kf . m . i)
In order to have the lowest freezing T° of solution, we need to know which solution has the highest value for the product (Kf . m . i)
Kf is a constant, so stays the same and m stays also the same because we have the same moles, in the same amount of solvent. In conclussion, same molality to all.
i defines everything. The i refers to the Van't Hoff factor which are the number of ions dissolved in solution. We assume 100 & of ionization so:
a. Glucose → i = 1
Glucose is non electrolytic, no ions formed
b. MgF₂ → Mg²⁺ + 2F⁻
i = 3. 1 mol of magnessium cation and 2 fluorides.
c. KBr → K⁺ + Br⁻
i = 2. 1 mol potassium cation and 1 mol of bromide anion
d. AlCl₃ → Al³⁺ + 3Cl⁻
i = 4. 1 mol of aluminum cation and 3 mol of chlorides.
Kf . m . 4 → option d will has the highest product, therefore will be the lowest freezing point.
Explanation:
here's the answer to the question
Answer:
a) Li2CO3
b) NaCLO4
c) Ba(OH)2
d) (NH4)2CO3
e) H2SO4
f) Ca(CH3COO)2
g) Mg3(PO4)2
f) Na2SO3
Explanation:
a) 2Li + CO3 ↔ Li2CO3
b) NaOH * HCLO4 ↔ NaCLO4 + H2O
c) Ba + 2H2O ↔ Ba(OH)2 +
d) 2NH4 + H2CO3 ↔ (NH4)2CO3 + H2O
c) SO2 + NO2 +H2O ↔ H2SO4 + NOx
f) 2CH3COOH + CaO ↔ Ca(CH3COOH)2 + H2O
g) 3MgO + 2H3PO4 ↔ Mg3(PO4)2 + H2O
h) NaOH + H2SO3 ↔ Na2SO3 + H2O
