Question

The measured osmotic pressure of seawater against pure water is 25 atm at 0°C. What is the concentration of salt in the seawater (assuming salt is the primary cause of the osmosis)?

a) 0.25 M
b) 1.25 M
c) 2.50 M
d) 5.00 M

Answer 1

The concentration of salt in seawater, calculated by using the osmotic pressure of 25 atm at 0°C, is approximately 2.50 M. This takes into account that NaCl dissociates into two particles, doubling the osmotic effect.

Step-by-step explanation:

The question asks us to calculate the concentration of salt in seawater by using the measured osmotic pressure. Osmotic pressure is a colligative property which depends on the number of solute particles in a solution. Sodium chloride (NaCl) in seawater dissociates into Na+ and Cl- ions, effectively doubling the number of particles.

Based on the Van't Hoff factor for NaCl, which is approximately 2 (since NaCl dissociates into 2 particles), we can use the formula for osmotic pressure:

\(\Pi = iMRT\)

Where \(\Pi\) is the osmotic pressure, i is the Van't Hoff factor, 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. Given that the osmotic pressure is 25 atm at 0°C (273 K), we can rearrange the equation to solve for M:

\(M = \frac{\Pi}{iRT}\)

Substituting in the given values: \(M = \frac{25 atm}{2 * 0.0821 L·atm/(K·mol) * 273 K}\)

When calculated, M is found to be greater than the option a) 0.25 M and closest to option c) 2.50 M. Therefore, the concentration of salt in the seawater is approximately 2.50 M.