Given that the density of Hg(l) at 0°C is about 14 g mL-1, which of the following is closest to the — Stoichiometry Chemistry Question

STEPS:
1. Identify the formula relating density, mass, and volume: Density ($D$) is defined as mass ($m$) divided by volume ($V$):
$$D = \frac{m}{V}$$
Rearranging this formula to solve for volume yields:
$$V = \frac{m}{D}$$
2. Find the molar mass of Mercury (Hg): Liquid mercury ($\text{Hg}(l)$) has a molar mass of approximately $200.59\text{ g mol}^{-1}$, which can be found on the periodic table included in the exam booklet. For a quick mental estimation on Section I—where calculators are not permitted—this value can be rounded to $200.\text{ g mol}^{-1}$.
3. Calculate the mass of exactly one mole of $\text{Hg}(l)$:
$$\text{Mass of 1 mole of Hg} = 1\text{ mol} \times 200.\text{ g mol}^{-1} = 200.\text{ g}$$
4. Calculate the volume of one mole of $\text{Hg}(l)$: Substitute the estimated mass ($200.\text{ g}$) and the given density ($14\text{ g mL}^{-1}$) into our rearranged density equation:
$$V = \frac{200.\text{ g}}{14\text{ g mL}^{-1}} = \frac{100}{7}\text{ mL} \approx \mathbf{14.3\text{ mL}}$$
This estimated volume is closest to 14 mL, which corresponds to Option D.

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WHY_OTHERS_WRONG:
* A is incorrect: $0.070\text{ mL}$ is obtained if a student mistakenly inverts the volume calculation and divides the density by the mass ($\frac{14}{200} = 0.070$) instead of dividing the mass by the density.
* B is incorrect: $0.14\text{ mL}$ represents a decimal place-value error during division, or is the result of using an incorrect value for the molar mass of mercury (such as its atomic number, 80).
* C is incorrect: $1.4\text{ mL}$ is off by a factor of 10, which typically occurs due to a decimal-shifting mistake during long division (e.g., incorrectly writing the decimal point when dividing 200 by 14).
* E is incorrect: $28\text{ mL}$ is double the correct volume, which could occur if a student mistakenly assumed that mercury exists as a diatomic molecule (such as $\text{Hg}_2$) under these conditions or made a multiplication error during division.