Question

Calculate the osmotic pressure of a solution containing 19.65 mg of hemoglobin in 14.2 mL of solution at 17 ∘C . The molar mass of hemoglobin is 6.5×104 g/mol.

Answer 1

Answer on Question # 73309 - Chemistry - General Chemistry

Calculate the osmotic pressure of a solution containing $19.65\mathrm{mg}$ of hemoglobin in $14.2\mathrm{mL}$ of solution at $17^{\circ}\mathrm{C}$. The molar mass of hemoglobin is $6.5\times 10^{4}\mathrm{g/mol}$.

Solution

In order to calculate the osmotic pressure of non-electrolytes, we use the van't Hoff Equation:

$$\Pi = \mathrm{MRT},$$

where $\Pi$ is the osmotic pressure in atm, M is the molar concentration in mol/L, R is the universal gas constant and T is the absolute temperature.

The absolute temperature is:

$$T = {}^{\circ}C + 273 = 17 + 273 = 290\ \mathrm{K}.$$

The amount of substance of hemoglobin is:

$$n = 19.65\ \mathrm{mg} \left(1\ \mathrm{g}/1000\ \mathrm{mg}\right) \left(1\ \mathrm{mol}/6.5 \times 10^{4}\ \mathrm{g}\right) = 3.023 \times 10^{-7}\ \mathrm{mol}.$$

The molarity of solution is:

$$M = \frac{3.023 \times 10^{-7}\ \mathrm{mol}}{0.0142\ \mathrm{L}} = 2.13 \times 10^{-5}\ \mathrm{mol/L}.$$

The osmotic pressure of the solution is:

$$\Pi = \frac{2.13 \times 10^{-5}\ \mathrm{mol/L} \cdot 0.08205\ \mathrm{L} \cdot \mathrm{atm/mol} \cdot \mathrm{K}}{290\ \mathrm{K}} = 5.07 \times 10^{-4}\ \mathrm{atm},\ \text{or}\ 51\ \mathrm{Pa}.$$

Answer: $5.07 \times 10^{-4}$ atm, or 51 Pa.

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