osmotic-pressure

To calculate the osmotic pressure of the solution at the new temperature, we can use the formula:

\( \pi = \text{CRT} \)

Since the concentration \( C \) and the gas constant \( R \) remain constant, the ratio of the osmotic pressures at two different temperatures is given by:

\( \frac{\pi_1}{\pi_2} = \frac{T_1}{T_2} \)

Thus, we can express the osmotic pressure at the second temperature as:

\( \pi_2 = \frac{\pi_1 \cdot T_2}{T_1} \)

Substituting the given values:

\( \pi_2 = \frac{7 \times 10^5 \times 283}{273} \)

After calculation, this simplifies to:

\( \pi_2 = 72.56 \times 10^4 \, \text{Nm}^{-2} \)

An artificial cell contains a 0.2 M glucose solution within a semipermeable membrane. When this cell is placed in a 0.05 M NaCl solution at 300 K, we need to determine the osmotic pressure developed, expressed as × 10⁻¹ bar.

Given:

Ideal Gas Constant, R = 0.083 L bar mol⁻¹ K⁻¹

Assume complete dissociation of NaCl.

Explanation

The osmotic pressure (\(\pi\)) is calculated using the formula:

\( \pi = \left( i_1 C_1 - i_2 C_2 \right) R T \)

where:

\(i_1\) and \(C_1\) are the van 't Hoff factor and concentration for glucose.

\(i_2\) and \(C_2\) are the van 't Hoff factor and concentration for NaCl.

\(R\) is the ideal gas constant.

\(T\) is the temperature in Kelvin.

For glucose:

\(i_1 = 1\) (glucose does not dissociate)

\(C_1 = 0.2\) M

For NaCl:

\(i_2 = 2\) (NaCl dissociates into two ions, Na⁺ and Cl⁻)

\(C_2 = 0.05\) M

Plugging in the values:

\( \pi = \left(1 \times 0.2 - 2 \times 0.05\right) \times 0.083 \times 300 \)

Calculating:

\( \pi = (0.2 - 0.1) \times 0.083 \times 300 \)

\( \pi = 0.1 \times 0.083 \times 300 \)

\( \pi = 2.5 \text{ bar} \)

Hence, the osmotic pressure developed is \(25 \times 10^{-1}\) bar.

The concept we are utilizing to solve this problem is osmotic pressure. Osmotic pressure is the pressure required to stop the flow of solvent into a solution through a semipermeable membrane. For dilute solutions, osmotic pressure behaves similarly to an ideal gas, hence the formula we use is similar to the ideal gas law:

\(\Pi = \frac{n}{V}RT\)

where:

• \(\Pi\) is the osmotic pressure,
• \(n\) is the number of moles of solute,
• \(V\) is the volume of the solution,
• \(R\) is the ideal gas constant, and
• \(T\) is the temperature in Kelvin.

First, we convert all given quantities to the appropriate units.

• \(\Pi = 1.29\) mbar is equivalent to \(1.29 \times 10^{-3}\) bar (since 1 bar = 1000 mbar),
• \(V = 300\) cm³ is equivalent to \(0.3\) L (since 1 L = 1000 cm³),
• \(T = 300\) K,
• \(R = 0.083\) L bar K\(^{-1}\) mol\(^{-1}\).

Instead of the number of moles, we are given the mass of the solute. However, we can replace the moles with mass using the relationship \(n = \frac{m}{M}\), where \(M\) is the molar mass, and \(m\) is the mass of the solute.

So the equation becomes:

\(\Pi = \frac{m}{MV}RT\)

We are trying to solve for \(M\), so rearrange the equation to isolate \(M\):

\(M = \frac{mRT}{\Pi V}\)

Finally, substitute the given values into the equation:

\(M = \frac{0.63 \times 0.083 \times 300}{1.29 \times 10^{-3} \times 0.3} \approx 40535 \, \text{g/mol}\)

So the molar mass of the protein is approximately 40535 g/mol.