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Chapter 13: Problem 66

Use the concentration of an isotonic saline solution, \(0.92 \% \mathrm{NaCl}(\mathrm{mass} / \text { volume }),\) to determine the osmotic pressure of blood at body temperature, \(37.0^{\circ} \mathrm{C}\). [Hint: Assume that \(\mathrm{NaCl}\) is completely dissociated in aqueous solutions.]

Short Answer

Expert verified
The osmotic pressure of blood at body temperature is approximately 7.98 atm.

Step by step solution

01

Understanding the problem and gathering information

Given concentration of NaCl is 0.92%, which means 0.92 g of NaCl are present in 100 g of solution. As this is a mass/volume percent, and density of water is approximately 1 g/cm^3 or 1 g/ml, we can consider this as 0.92 grams in 100 ml solution. Temperature is given as 37.0 °C, which in Kelvin (K) is \(37.0 + 273.15 = 310.15 K\) (since absolute temperature is required for calculations involving gas laws). Following the hint, we know that NaCL dissociates completely in solution into its ions, Na+ and Cl-, making the Van't Hoff factor (i) 2.
02

Conversion of concentration

We first convert the concentration of NaCl from g/100ml to moles/liter (M), since osmotic pressure is typically expressed in these units. We use the molar mass of NaCl, which is approximately 58.44 g/mol. \( \text {Convert mass of NaCl to moles} = \frac{0.92 \text{ g NaCl}}{58.44 \text{ g/mol NaCl}} = 0.0157 \text{ moles NaCl} \). Next, convert the volume from milliliters to liters: \(0.1 \text { liters} \). Thus, the molar concentration is \( \frac{0.0157 \text{ moles}}{0.1 \text{ liters}} = 0.157 \text{ M NaCl} \)
03

Calculation of osmotic pressure

We then use the formula for osmotic pressure: \( \text{Osmotic pressure} = i \cdot n/V \cdot R \cdot T\) where \(n/V\) is the molar concentration (in moles/liter), R is the gas constant (0.0821 liter·atm / (mol·K)), and T is the absolute temperature. Substituting the given values: \( \text{Osmotic pressure} = 2 \cdot 0.157 \text { M} \cdot 0.0821 \text { liter·atm / (mol·K)} \cdot 310.15 \text { K} = 7.98 \text { atm} \).

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Most popular questions from this chapter

The concentration of \(\mathrm{N}_{2}\) in the ocean at \(25^{\circ} \mathrm{C}\) is \(445 \mu \mathrm{M} .\) The Henry's law constant for \(\mathrm{N}_{2}\) is \(0.61 \times 10^{-3} \mathrm{mol} \mathrm{L}^{-1} \mathrm{atm}^{-1} .\) Calculate the mass of \(\mathrm{N}_{2}\) in a liter of ocean water. Calculate the partial pressure of \(\mathrm{N}_{2}\) in the atmosphere.

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A saturated solution prepared at \(70^{\circ} \mathrm{C}\) contains \(32.0 \mathrm{g}\) CuSO \(_{4}\) per 100.0 g solution. A 335 g sample of this solution is then cooled to \(0^{\circ} \mathrm{C}\) and \(\mathrm{CuSO}_{4} \cdot 5 \mathrm{H}_{2} \mathrm{O}\) crystallizes out. If the concentration of a saturated solution at \(0^{\circ} \mathrm{C}\) is \(12.5 \mathrm{g} \mathrm{CuSO}_{4} / 100 \mathrm{g}\) soln, what mass of \(\mathrm{CuSO}_{4} \cdot 5 \mathrm{H}_{2} \mathrm{O}\) would be obtained? [Hint: Note that the solution composition is stated in terms of \(\mathrm{CuSO}_{4}\) and that the solid that crystallizes is the hydrate \(\left.\mathrm{CuSO}_{4} \cdot 5 \mathrm{H}_{2} \mathrm{O} .\right]\)

Calculate the mole fraction of solute in the following aqueous solutions: (a) \(21.7 \% \mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{OH},\) by mass; (b) \(0.684 m \mathrm{CO}\left(\mathrm{NH}_{2}\right)_{2}\) (urea).
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