Problem 74 - Chemistry: The Central Science

Question

Seawater contains \(3.4 \mathrm{~g}\) of salts for every liter of solution. Assuming that the solute consists entirely of \(\mathrm{NaCl}\) (over \(90 \%\) is), calculate the osmotic pressure of seawater at \(20^{\circ} \mathrm{C}\).

Step-by-Step Solution

Answer

The osmotic pressure of seawater at \(20^{\circ} \mathrm{C}\) is approximately \(1.41 \mathrm{atm}\).

Step 1: Identify the variables and constants

Given variables: - Salt concentration in seawater: \(3.4 \mathrm{g/L}\) - Temperature: \(20^{\circ} \mathrm{C}\) - The molar mass of NaCl: \(58.44 \mathrm{g/mol}\) Constants: - Gas constant (\(R\)): \(0.0821 \mathrm{L \cdot atm / (mol \cdot K)}\)

Step 2: Convert temperature to Kelvin

The temperature must be in Kelvin (K) to use the gas constant. Convert the given Celsius temperature to Kelvin: Temperature, \(T = 20^{\circ} \mathrm{C} + 273.15 = 293.15 \mathrm{K}\)

Step 3: Convert mass concentration to molar concentration

Calculate the molar concentration, \(\mathrm{M}\), using the given mass concentration and the molar mass of NaCl: \(\mathrm{M} = \frac{3.4 \mathrm{g/L}}{58.44 \mathrm{g/mol}} = 0.0582 \mathrm{mol/L}\)

Step 4: Calculate the osmotic pressure using the formula

Use the formula for osmotic pressure derived from the ideal gas law: Osmotic pressure, \(Π = \mathrm{M} \cdot R \cdot T\) Plug in the values for \(\mathrm{M}, R,\) and \(T\): \( Π = (0.0582 \mathrm{mol/L}) \cdot (0.0821 \mathrm{L \cdot atm / (mol \cdot K)}) \cdot (293.15 \mathrm{K}) \)

Step 5: Calculate the osmotic pressure

Solve for the osmotic pressure, \(Π\): \( Π \approx 1.41 \mathrm{atm} \) Therefore, the osmotic pressure of seawater at \(20^{\circ} \mathrm{C}\) is approximately \(1.41 \mathrm{atm}\).