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Among the following 1 molal aqueous solutions of cobalt complexes, which will show the highest freezing point (least depression)?
The depression in freezing point $\Delta T_f = i \cdot K_f \cdot m$, where $i$ = van't Hoff factor (number of particles from ionization).
The compound with the fewest ions in solution gives the smallest $\Delta T_f$ and therefore the highest freezing point.
- $[\text{Co(H}_2\text{O)}_4\text{Cl}_2]\text{Cl}$: ionizes to give 2 particles → $i = 2$
- $[\text{Co(H}_2\text{O)}_3\text{Cl}_3]$: non-electrolyte → $i = 1$ (no ions) — smallest depression
- $[\text{Co(H}_2\text{O)}_6]\text{Cl}_3$: gives 4 particles → $i = 4$
- $[\text{Co(H}_2\text{O)}_5\text{Cl}]\text{Cl}_2$: gives 3 particles → $i = 3$
Highest freezing point → smallest $i$ → Option B ($i=1$).
How many moles of NaCl are produced from 16.5 g HCl? \(\mathrm{HCl + NaOH \rightarrow NaCl + H_2O}\)
Molar mass HCl = 36.5 g/mol. Moles HCl = 16.5/36.5 = 0.452 mol. 1:1 ratio, so moles NaCl = 0.452 mol.
Draft loss for year: $320. Adjustments needed: (1) Irrecoverable debts of $930 recovered; (2) Existing allowance $520, required allowance = 5% of trade receivables $9,600. What is the revised profit or loss?
Option D (Profit of $650) is correct.
Recovery of debt: +$930
Required allowance = 5% × $9,600 = $480; Reduction = $520 − $480 = +$40
Revised = −$320 + $930 + $40 = +$650 profit
- Across a period: decreases left to right (increasing Z*)
- Down a group: increases (new shells added)
So increasing order: Cl < P < Mg < Ca
A person throws balls vertically upward one after another at intervals of 2 s, all with the same speed. For more than two balls to be in the air simultaneously, the throwing speed must be:
Time of flight for one ball $= \dfrac{2u}{g}$.
For more than 2 balls in the air, a third ball is thrown before the first lands, so time of flight $> 4\text{ s}$ (two intervals of 2 s each):
$\dfrac{2u}{g} > 4 \Rightarrow u > 2g = 2 \times 9.8 = 19.6\text{ m/s}$
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