A 0.0020 m aqueous solution of an ionic compound Co(NH3)5(NO2)CI freez...
Given:
Molality (m) = 0.0020 m
Freezing point depression (ΔTf) = -0.007320C
Kf = 1.860C/m
To find:
Number of moles of ions produced by 1 mol of Co(NH3)5(NO2)CI
Solution:
We know that for an ionic compound, the number of moles of ions produced is equal to the sum of the moles of cations and anions.
Let x be the number of moles of Co(NH3)5(NO2)CI in the solution.
Molar mass of Co(NH3)5(NO2)CI = (1 x molar mass of Co) + (5 x molar mass of NH3) + (1 x molar mass of NO2) + (1 x molar mass of Cl)
= (1 x 58.93 g/mol) + (5 x 17.03 g/mol) + (1 x 46.01 g/mol) + (1 x 35.45 g/mol)
= 287.23 g/mol
Mass of Co(NH3)5(NO2)CI in 1 L of solution = 0.0020 mol x 287.23 g/mol = 0.574 g
Now, we can use the freezing point depression formula:
ΔTf = Kf x m x i
where i is the van't Hoff factor, which represents the number of particles (ions, molecules, or formula units) produced by 1 mol of the solute.
Rearranging the formula, we get:
i = ΔTf / (Kf x m)
Substituting the given values, we get:
i = -0.007320C / (1.860C/m x 0.0020 m) = 1.97
Since i must be a whole number, we round it to the nearest integer, which is 2.
Therefore, 1 mol of Co(NH3)5(NO2)CI produces 2 moles of ions.
Answer: Option B (2)
A 0.0020 m aqueous solution of an ionic compound Co(NH3)5(NO2)CI freez...
Given:
molality = 0.002molal
Tf = -0.00732•C
Tf• = 0•C
∆Tf = 0•C-(-0.00732•C)=0.00732•C
∆Tf = i × Kf ×m
i = = ∆Tf/(Kf ×m)
i =0.00732•C/(1.86•C/m × 0.002m)
i =1.9677 ~ 2