Final answer:
The molarity of ocean water concerning salt ions is calculated as 35 M, considering 35 moles of salt per liter. Using the dissociation of NaCl into two ions, the total ionic concentration becomes 70 M. The osmotic pressure at 25°C, calculated using the ideal gas law equation, is 1717.98 atm, which is theoretically significant compared to real-world conditions.
Step-by-step explanation:
The given question involves calculating the molarity of a salt solution in ocean water and the resulting osmotic pressure when compared to pure water at 25°C. To begin, we identify from the question that ocean water contains approximately 35 moles of salt ions per every 1000 moles of water. As ocean water has a similar density to pure water, we can use a volume of 1 liter as an approximation for 1000 moles of water (since the molar mass of water is approximately 18 g/mol).
Given the density of water is about 1 kg/L, and using the concept that 1 mol of NaCl dissociates into 2 moles of ions (1 mole of Na+ and 1 mole of Cl-), we can calculate the molarity and subsequently, the osmotic pressure using the formula:
II = MRT
where II is the osmotic pressure, M is the molarity of the solution, R is the ideal gas constant (0.0821 L·atm/K·mol), and T is the temperature in Kelvin (298 K for 25°C). First, we convert the number of moles of salt to molarity:
Molarity (M) = Moles of salt / Volume of solution in liters
Since 1 liter of ocean water contains 35 moles of salt ions, the molarity equates to:
M = 35 moles / 1 L = 35 M
Considering that each mole of NaCl produces 2 moles of ions:
Total concentration of ions = 2 × 35 M = 70 M
Plugging this into the equation for osmotic pressure:
II = (70 M) (0.0821 L·atm/K·mol) (298 K)
II = 1717.98 atm
This value indicates the osmotic pressure between ocean water and pure water at 25°C is approximately 1718 atm, which is a theoretical value and may differ in reality due to assumptions like ideal behaviour and ignoring the presence of other ions and the actual density differences of seawater.