Problem 31 - Chemistry: An Atoms-Focused Approach

Question

After \(100.0 \mathrm{mL}\), of a solution of physiological saline \((0.90 \%\) NaCl by mass) is diluted by the addition of \(250.0 \mathrm{mL}\) of water, what is the osmotic pressure of the final solution at \(37^{*} \mathrm{C} ?\) Assume that \(\mathrm{NaCl}\) dissociates completely into \(\mathrm{Na}^{+}(a q)\) and \(\mathrm{Cl}^{+}(a q)\).

Step-by-Step Solution

Answer

Answer: The osmotic pressure of the final solution after dilution is approximately 2.31 atm.

Step 1: Calculate the mass of NaCl in the initial solution

The initial solution is physiological saline with 0.90% NaCl by mass. We have 100.0 mL of this solution. Let's find the mass of NaCl in 100.0 mL of physiological saline. The mass of NaCl in solution = mass percentage × volume of solution × density of solution We are not given the density of the physiological saline solution. However, we can assume it to be approximately equal to the density of water, 1 g/mL, since the amount of solute is quite small. mass of NaCl = (0.90% × 100.0 mL × 1g/mL) = 0.009 × 100 g = 0.9 g

Step 2: Calculate the moles of NaCl in the initial solution

First we need to find the molar mass of NaCl. From the periodic table, the atomic mass of Na is 22.99 g/mol, and Cl is 35.45 g/mol. Thus, the molar mass of NaCl = 22.99 + 35.45 = 58.44 g/mol. Now, we will find the moles of NaCl in the initial solution: moles = mass / molar mass → moles of NaCl = 0.9 g / 58.44 g/mol ≈ 0.0154 moles

Step 3: Calculate the number of particles formed by NaCl dissociation

NaCl dissociates completely into Na+ and Cl- ions. So upon dissociation, the number of particles formed by one molecule of NaCl is 2 (1 Na+ and 1 Cl-). Therefore, total moles of particles in solution = moles of NaCl × particles formed per NaCl = 0.0154 moles × 2 = 0.0308 moles

Step 4: Calculate the final volume of solution

We are adding 250.0 mL of water to the 100.0 mL of the initial saline solution. So the final volume of the solution is: Final volume = initial volume + volume of water added = 100.0 mL + 250.0 mL = 350 mL

Step 5: Calculate the final concentration of particles

Now we need to find the concentration of particles in the final solution. Concentration is calculated as moles of particles divided by the volume of the solution in liters. Final concentration = (moles of particles) / (final volume in liters) Final concentration = 0.0308 moles / 0.35 L ≈ 0.0880 M

Step 6: Convert the temperature to Kelvin

The given temperature is 37°C. To convert it to Kelvin, we add 273.15: T(K) = T(°C) + 273.15 → T(K) = 37°C + 273.15 ≈ 310.15 K

Step 7: Calculate the osmotic pressure

Now we can use the osmotic pressure formula: osmotic pressure (Π) = nRT/V Here, n is the moles of solute particles, R is the ideal gas constant (0.0821 L atm / K mol), T is the temperature in Kelvin, and V is the volume in liters. Π = (0.0308 moles × 0.0821 L atm / K mol × 310.15 K) / 0.35 L ≈ 2.31 atm The osmotic pressure of the final solution after dilution is approximately 2.31 atm.

Key Concepts

NaCl dissociation

Understanding the dissociation of sodium chloride (NaCl) is key in chemistry when dealing with solutions. NaCl is an ionic compound, which means it is composed of positively charged sodium ions (\( \mathrm{Na}^{+} \)) and negatively charged chloride ions (\( \mathrm{Cl}^{-} \)). When NaCl dissolves in water, it dissociates completely into these ions.

This complete dissociation is essential because it determines how substances like NaCl contribute to the properties of a solution. For every molecule of NaCl, it will form one \( \mathrm{Na}^{+} \) and one \( \mathrm{Cl}^{-} \), doubling the number of particles in the solution compared to the moles of NaCl originally present. This process is crucial for our calculations related to moles, concentration, and osmotic pressure.

In our exercise, understanding that NaCl dissociates completely into 2 ions per formula unit helps in correctly calculating the number of particles present in our solution.

moles calculation

Calculating moles is a fundamental step to understanding and solving chemical problems. When determining the moles of a substance, we use the formula: \[ \text{moles} = \frac{\text{mass of substance (g)}}{\text{molar mass (g/mol)}} \]

For example, in the question at hand, we calculated the mass of NaCl to be 0.9 grams. The molar mass of NaCl is the sum of the atomic masses of sodium (Na) and chlorine (Cl), which are approximately 22.99 g/mol and 35.45 g/mol, respectively. Thus, the molar mass of NaCl is 58.44 g/mol.

Using these values, the moles of NaCl can be found:

• Mass of NaCl = 0.9 g
• Molar Mass of NaCl = 58.44 g/mol
• Moles of NaCl = 0.9 g / 58.44 g/mol ≈ 0.0154 moles

This calculation is crucial as it serves as the basis for determining the concentration and later the osmotic pressure of the solution.

concentration calculation

To calculate concentration, we determine how many moles of particles are present in a given volume of solution. Concentration is expressed in molarity (M), which is moles per liter (mol/L).

In our problem, after calculating the moles of NaCl, we considered its dissociation into ions. Since each NaCl formula unit results in two ions (\( \mathrm{Na}^{+} \) and \( \mathrm{Cl}^{-} \)), we double the moles to find the total moles of particles:

• Moles of NaCl = 0.0154 moles
• Total moles of particles = 0.0154 moles × 2 = 0.0308 moles

The next step is to use the total volume of the solution, which in this case is the sum of the initial solution's volume and the water added (100.0 mL + 250.0 mL = 350.0 mL = 0.35 L).

The concentration of particles in the final solution is:\[ \text{Concentration} = \frac{\text{Total moles of particles}}{\text{Total volume in liters}} = \frac{0.0308 \text{ moles}}{0.35 \text{ L}} \approx 0.0880 \text{ M} \]

Understanding concentration is important for further calculations such as osmotic pressure.