A sample of drinking water was found to be severely contaminated with chloroform, CHCl3, supposed to be carcinogenic in nature. The level of contamination was 15 ppm (by mass). Determine the molality of chloroform in the water sample.

The correct option is 1.25 x 10-4 m (which represents $1.25\times {10}^{-4}m$).

Step 1: Understand the given data
The concentration of chloroform (${\mathrm{CHCl}}_3$) in water is given as $15\text{ppm}$ (parts per million) by mass.
By definition, $15\text{ppm}$ by mass means that there are $15\text{grams}$ of chloroform present in ${10}^6\text{grams}$ of the solution.
Therefore:
Mass of solute (chloroform, $W_B$) = $15\text{g}$
Mass of solution = ${10}^6\text{g}$

Step 2: Determine the mass of the solvent (water)
Since the mass of the solute ($15\text{g}$) is extremely small compared to the mass of the solution (${10}^6\text{g}$), the mass of the solvent (water, $W_A$) can be approximated as the mass of the solution itself:
Mass of solvent ($W_A$) ≈ ${10}^6\text{g}$ = ${10}^3\text{kg}$

Step 3: Calculate the molar mass of chloroform (${\mathrm{CHCl}}_3$)
The molar mass ($M_B$) of chloroform is calculated by summing the atomic masses of its constituent atoms:
$M_B=\text{Atomic mass of C}+\text{Atomic mass of H}+3\times \left(\text{Atomic mass of Cl}\right)$
$M_B=12+1+3\times 35.5$
$M_B=13+106.5=119.5\text{g/mol}$

Step 4: Calculate the molality (m) of the solution
Molality is defined as the number of moles of solute dissolved per kilogram of the solvent:
$\text{Molality (m)}=\frac{\text{Moles of solute}}{\text{Mass of solvent in kg}}=\frac{W_B}{M_B\times W_A\text{(in kg)}}$
Substituting the values into the formula:
$m=\frac{15}{119.5\times {10}^3}$
$m\approx 0.1255\times {10}^{-3}\text{m}$
$m\approx 1.25\times {10}^{-4}\text{m}$

This derivation leads directly to the correct option of 1.25 x 10-4 m.