Percent by mass tells you what portion of the total mass is ethanol. To find the mass of ethanol, multiply the total sample mass by the fraction that is ethanol: 0.89 × 200 g = 178 g. So the ethanol mass in the sample is 178 g. The other numbers would correspond to smaller fractions (for example, 22 g is about 11% of 200 g, 100 g is 50%, and 46 g is about 23%), which don’t match the given 89% by mass.

This question tests how adding a nonvolatile solute affects the freezing point. Dissolving particles in a solvent lowers the solvent’s mole fraction and lowers its vapor pressure, which shifts the equilibrium between the solid and liquid to require a lower temperature for freezing. In other words, the solution’s liquid phase remains stable down to a lower temperature because the dissolved particles disrupt the orderly formation of the solid. This freezing-point depression is a classic colligative property and scales with the number of dissolved particles (ΔTf = iKf m for ideal solutions). So, adding a nonvolatile solute causes the freezing point to decrease.

Colligative properties depend on the number of dissolved particles, not on their identity. The size of both boiling point elevation and freezing point depression is set by how many solute particles are in solution, which is captured by molality and the van't Hoff factor. The equations ΔTb = i Kb m and ΔTf = i Kf m show this directly: m is the molality (moles of solute per kilogram of solvent), and i accounts for how many particles the solute produces in solution (dissociation or association). For a non-electrolyte, i is about 1; for a salt like NaCl, i is about 2 because it dissociates into two ions, increasing the shifts. Increasing concentration or a higher degree of dissociation makes the changes larger. Time isn’t a factor in the magnitude once equilibrium is reached, and the molar mass of the solute doesn’t directly influence these effects because what matters is particle count, not mass. Solvent polarity can influence how well something dissolves, but after dissolution the shifts depend on the number of particles.

In first-order decay, activity falls exponentially with time, and the fraction remaining ties directly to time via N/N0 = (1/2)^{t/t1/2}. Here, we want the activity to be 20% of its initial value, so 0.20 = (1/2)^{t/14.3}. Taking natural logs gives t/14.3 = ln(0.20)/ln(0.50). Using ln(0.20) ≈ -1.6094 and ln(0.50) ≈ -0.6931, we get t ≈ 14.3 × (2.322) ≈ 33.2 days. So about 33 days. Intuitively, two half-lives (about 28.6 days) would leave 25% remaining; reaching 20% requires a bit more time, which matches a little over 33 days. The other time options correspond to significantly more or fewer half-lives and would produce fractions well below 20%.