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 ($i=1$).

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.

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)

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):

$\dfrac{2u}{g} > 4 \Rightarrow u > 2g = 2 \times 9.8 = 19.6\text{ m/s}$