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Chapter 13: Problem 64

The molecular mass of hemoglobin is \(6.86 \times 10^{4} \mathrm{u}\) What mass of hemoglobin must be present per \(100.0 \mathrm{mL}\) of a solution to exert an osmotic pressure of \(7.25 \mathrm{mmHg}\) at \(25^{\circ} \mathrm{C} ?\)

Short Answer

Expert verified
Approximately 2.67 grams of hemoglobin must be present in 100.0 mL of the solution to exert the given osmotic pressure at 25 degrees Celsius.

Step by step solution

01

Understand the Relation Between Osmotic Pressure and Concentration

According to the ideal gas law and the concept of osmotic pressure, the osmotic pressure \( P \) of a solution is given by \( P = C R T \), where \( c \) is the molar concentration of the solute, \( R \) is the ideal gas constant, and \( T \) is the absolute temperature. Here, we need to find \( c \), knowing that \( P = 7.25 \, mmHg \), \( R = 0.0821 \, L.atm/K.mol \) (in appropriate units), and \( T = 25^{\circ}C = 298 \, K \) (converting from Celsius to K).
02

Convert pressure to appropriate units

Before we directly put the values into our equation, it is important that the units match. Therefore, convert pressure from mmHg to atm. Knowing that 1 atmosphere is equivalent to 760 mmHg, the pressure in atmospheres is \( P = 7.25 \, mmHg \times \frac{1 \, atm}{760 \, mmHg} = 0.00954 \, atm \).
03

Calculate the concentration

Now, rearrange the ideal gas law to solve for the molar concentration \( c \). After rearranging, we get \( c = \frac{P}{RT} \). Substituting our known values, we find \( c = \frac{0.00954 \, atm}{0.0821 \, L.atm/K.mol \times 298 \, K} = 0.00039 \, moles/L \). This is the molar concentration of hemoglobin that we need.
04

Calculate the mass of Hemoglobin

We know that hemoglobin has a molar mass of \(6.86 \times 10^{4} \, u\) or \(6.86 \times 10^{4} \, g/mol\). To find out the mass of hemoglobin that must be present to produce this concentration in 100.0 mL of solution, we need to multiply the molar concentration by the volume (in liters) and then by the molar mass. Doing so gives us \(0.00039 \, moles/L \times 0.1 \, L \times 6.86 \times 10^{4} \, g/mol = 2.67 \, g\). So, we need about 2.67 grams of hemoglobin.

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Most popular questions from this chapter

What are the partial and total vapor pressures of a solution obtained by mixing 35.8 g benzene, \(\mathrm{C}_{6} \mathrm{H}_{6}\) and \(56.7 \mathrm{g}\) toluene, \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CH}_{3},\) at \(25^{\circ} \mathrm{C} ? \mathrm{At} 25^{\circ} \mathrm{C}\) the vapor pressure of \(\mathrm{C}_{6} \mathrm{H}_{6}=95.1 \mathrm{mm} \mathrm{Hg} ;\) the vapor pressure of \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CH}_{3}=28.4 \mathrm{mmHg}\).

How many grams of iodine, \(I_{2}\), must be dissolved in \(725 \mathrm{mL}\) of carbon disulfide, \(\mathrm{CS}_{2}(d=1.261 \mathrm{g} / \mathrm{mL}),\) to produce a \(0.236 \mathrm{m}\) solution?

Explain the observation that all metal nitrates are water soluble but many metal sulfides are not. Among metal sulfides, which would you expect to be most soluble?

The most likely of the following mixtures to be an ideal solution is (a) \(\mathrm{NaCl}-\mathrm{H}_{2} \mathrm{O} ;\) (b) \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{OH}-\mathrm{C}_{6} \mathrm{H}_{6}\) (c) \(\mathrm{C}_{7} \mathrm{H}_{16}-\mathrm{H}_{2} \mathrm{O} ;\) (d) \(\mathrm{C}_{7} \mathrm{H}_{16}-\mathrm{C}_{8} \mathrm{H}_{18}\)

Which aqueous solution from the column on the right has the property listed on the left? Explain your choices. $$\begin{array}{ll}\hline \text { Property } & \text { Solution } \\\\\hline \text { 1. lowest electrical } & \text { a. } 0.10 \mathrm{m} \mathrm{KCl}(\mathrm{aq}) \\\\\text { conductivity } & \\\\\text { 2. } \text { lowest boiling } & \text { b. } 0.15 \mathrm{m}\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}(\mathrm{aq}) \\\\\text { point } & \\\\\text { 3. highest vapor pressure } & \text { c. } 0.10 \mathrm{m} \mathrm{CH}_{3} \mathrm{COOH}(\mathrm{aq}) \\\\\text { of water at } 25^{\circ} \mathrm{C} & \\\\\text { 4. lowest freezing point } & \text { d. } 0.05 \mathrm{m} \mathrm{NaCl} \\\\\hline\end{array}$$
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