Final answer:
The density of water would increase by 1200% in the deepest ocean of about 12.0km. However, this calculation assumes the density of water is constant, which may not be the case in reality.
Step-by-step explanation:
In this question, we are asked to calculate the percent increase in the density of water in the deepest ocean, which is about 12.0km deep. The pressure of sea water increases by 1.0atm for each 10m increase in depth. To find the percent increase in density, we need to consider the change in pressure. Since the pressure increases by 1.0atm for every 10m deep, the pressure at a depth of 12.0km would be 1.0atm x (12.0km / 10m) = 12.0atm.
The density of water is defined as mass divided by volume. Assuming the mass of water remains constant, and knowing that pressure is inversely proportional to volume (according to Boyle's Law), we can determine the percent decrease in volume due to the increased pressure. If the bulk modulus of water is the same as water and is constant, pressure is related to volume by the equation ΔV/V = -ΔP/P. Therefore, the percent decrease in volume (and thus density) can be calculated as follows:
Percent decrease in volume = (ΔV/V) * 100 = (-ΔP/P) * 100 = (-12.0atm/1.0atm) * 100 = -1200%.
Thus, the percent increase in density would be 1200%. This means that the density of water would increase by 1200% in the deepest ocean. It is important to note that this calculation assumes the density of water is constant all the way down, but in reality, the density of water can vary with depth due to changes in temperature and salinity. The actual pressure at the deepest ocean would be greater than that calculated under the assumption of constant density.