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K$_2$HgI$_4$ is 40% ionized in aqueous solution. What is the van't Hoff factor ($i$)?
K$_2$HgI$_4$ dissociates as: $\text{K}_2\text{HgI}_4 \to 2\text{K}^+ + \text{HgI}_4^{2-}$ (3 particles total if fully ionized)
Degree of ionization $\alpha = 0.40$, total particles per formula unit if fully ionized = 3.
$i = 1 + (n-1)\alpha = 1 + (3-1)(0.40) = 1 + 0.80 = \mathbf{1.80}$
- $2$ moles of resin ($2 \times 206\text{ g}$) react with $1$ mole of $Ca^{2+}$.
- Total mass of resin = $412\text{ g}$.
- Uptake per gram = $\frac{1\text{ mole } Ca^{2+}}{412\text{ g resin}}$.
The volume of a cube is 64.
Compare:
Column A: The area of the base of the cube
Column B: $32$
For a cube with volume $V = s^3 = 64$, the side length is $s = \sqrt[3]{64} = 4$.
The base of a cube is a square, so its area $= s^2 = 4^2 = 16$.
Column A = 16, Column B = 32.
Since $16 < 32$, Column B is greater.
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