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Chapter 13: Problem 104

A small protein has a molar mass of \(1.50 \times 10^{4} \mathrm{~g} / \mathrm{mol}\). What is the osmotic pressure exerted at \(24.0^{\circ} \mathrm{C}\) by \(25.0 \mathrm{~mL}\) of an aqueous solution that contains \(37.5 \mathrm{mg}\) of the protein?

Short Answer

Expert verified
The osmotic pressure is approximately \(\pi = 2.44 \times 10^{-3}\) atm.

Step by step solution

01

Understand the relevant formula

The osmotic pressure (\( \pi \)) can be calculated using the van't Hoff equation: \[\pi = \frac{nRT}{V}\], where \(n\) is the number of moles of solute, \(R\) is the ideal gas constant (0.0821 L路atm/(K路mol)), \(T\) is the temperature in Kelvin, and \(V\) is the volume of the solution in liters.
02

Convert the given mass of protein to moles

First, convert the mass (\(m\)) of protein from milligrams to grams. \[m = 37.5 \ \text{mg} \times \frac{1 \text{g}}{1000 \ \text{mg}} = 0.0375 \ \text{g}\] Next, calculate the number of moles (\(n\)) using the molar mass (\(M\)): \[n = \frac{m}{M} = \frac{0.0375 \ \text{g}}{1.50 \times 10^4 \ \text{g/mol}} = 2.50 \times 10^{-6} \ \text{mol}\]
03

Convert temperature to Kelvin

Convert the temperature (\(T\)) from Celsius to Kelvin using the equation: \[T(K) = T(^{\circ}C) + 273.15\] So, \[T = 24.0 + 273.15 = 297.15 \ \text{K}\]
04

Convert the volume to liters

Convert the volume (\(V\)) from milliliters to liters: \[V = 25.0 \ \text{mL} \times \frac{1 \ \text{L}}{1000 \ \text{mL}} = 0.025 \ \text{L}\]
05

Calculate osmotic pressure

Insert the values into the van't Hoff equation: \[\pi = \frac{(2.50 \times 10^{-6} \ \text{mol})(0.0821 \ \text{L路atm/K路mol})(297.15 \ \text{K})}{0.025 \ \text{L}}\] Simplify the calculation: \[\pi = \frac{(2.50 \times 10^{-6})(0.0821)(297.15)}{0.025}\] \[\pi = \frac{(6.1042375 \times 10^{-5})}{0.025} = 2.44 \times 10^{-3} \ \text{atm}\]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

van't Hoff equation
The van't Hoff equation is central to calculating osmotic pressure. Understanding it can be key to solving various chemistry problems. The equation is written as: \[ \pi = \frac{nRT}{V} \] Here, \( \pi \) represents osmotic pressure. The term \( n \) is the number of moles of solute, \( R \) is the ideal gas constant (0.0821 L路atm/(K路mol)), \( T \) is temperature in Kelvin, and \( V \) is the volume of the solution in liters. Pretty straightforward, right? By inserting values into this formula, we can determine the osmotic pressure for any solution as long as we know the values for these variables.
molar mass
Molar mass plays a vital role in our calculations. It helps us convert the mass of the solute into moles, which we then use in the van't Hoff equation. Molar mass is simply the mass of one mole of a substance and is usually expressed in grams per mole (g/mol). To find the number of moles (\( n \)), we use the formula: \[ n = \frac{m}{M} \] where \( m \) is the mass of the substance and \( M \) is its molar mass. In our example, the protein has a molar mass of \( 1.50 \times 10^4 \ \text{g/mol} \). Knowing this, we could convert 37.5 mg of the protein to moles by following the previous steps.
ideal gas constant
The ideal gas constant (\( R \)) is crucial in the van't Hoff equation, but what exactly is it? It connects the amount of substance, temperature, volume, and pressure in gas-related calculations. The value of \( R \) we use specifically for these kinds of chemical calculations is 0.0821 L路atm/(K路mol). This constant remains unchanged, making it reliable in a wide range of chemistry problems, including the osmotic pressure calculation we're working on. By ensuring \( R \) is in the right units, we can seamlessly integrate it into our equations for accurate results.
temperature conversion
For osmotic pressure calculations, we need the temperature in Kelvin. Most given temperatures are in Celsius, so conversion is needed. The conversion formula is: \[ T(K) = T(^{\circ}C) + 273.15 \] So, if we're given a temperature like 24.0掳C, we convert it to Kelvin by simply adding 273.15. This means: \[ T = 24.0 + 273.15 = 297.15 \ \text{K} \] Simple, right? This conversion allows us to use the temperature in calculations directly since Kelvin is the standard unit in thermodynamic equations.
volume conversion
Volume conversions are often necessary to ensure our units are consistent in equations. For the van鈥檛 Hoff equation, the volume should be in liters. Often, volumes are provided in milliliters (mL), so we need to convert them. The conversion is: \[ V = \text{mL} \times \frac{1 \text{L}}{1000 \text{mL} } \] For example, converting 25.0 mL to liters is simple: \[ 25.0 mL \times \frac{1 \text {L} }{1000 \text{mL} } = 0.025 \text {L} \] This small step is crucial as it ensures our values are compatible with the ideal gas constant's units in the van鈥檛 Hoff equation.

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Most popular questions from this chapter

The total concentration of dissolved particles in blood is \(0.30 \mathrm{M}\). An intravenous (IV) solution must be isotonic with blood, which means it must have the same concentration. (a) To relieve dehydration, a patient is given \(100 . \mathrm{mL} / \mathrm{h}\) of \(\mathrm{IV}\) glucose \(\left(\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}\right)\) for \(2.5 \mathrm{~h}\). What mass (g) of glucose does she receive? (b) If isotonic saline \((\mathrm{NaCl})\) is used, what is the molarity of the solution? (c) If the patient is given \(150 . \mathrm{mL} / \mathrm{h}\) of IV saline for \(1.5 \mathrm{~h}\), how many grams of \(\mathrm{NaCl}\) does she receive?

What are the most important differences between the phase diagram of a pure solvent and the phase diagram of a solution of that solvent?

Although other solvents are available, dichloromethane \(\left(\mathrm{CH}_{2} \mathrm{Cl}_{2}\right)\) is still often used to "decaffeinate" drinks because the solubility of caffeine in \(\mathrm{CH}_{2} \mathrm{Cl}_{2}\) is 8.35 times that in water. (a) A 100.0 -mL sample of cola containing 10.0 mg of caffeine is extracted with \(60.0 \mathrm{~mL}\) of \(\mathrm{CH}_{2} \mathrm{Cl}_{2} .\) What mass of caffeine remains in the aqueous phase? (b) A second identical cola sample is extracted with two successive \(30.0-\mathrm{mL}\) portions of \(\mathrm{CH}_{2} \mathrm{Cl}_{2} .\) What mass of caffeine remains in the aqueous phase after each extraction? (c) Which approach extracts more caffeine?

Soft drinks are canned under 4 atm of \(\mathrm{CO}_{2}\) and release \(\mathrm{CO}_{2}\) when the can is opened. (a) How many moles of \(\mathrm{CO}_{2}\) are dissolved in \(355 \mathrm{~mL}\) of soda in a can before it is opened? (b) After the soda has gone flat? (c) What volume (in L) would the released \(\mathrm{CO}_{2}\) occupy at \(1.00 \mathrm{~atm}\) and \(25^{\circ} \mathrm{C}\left(k_{\mathrm{H}}\right.\) for \(\mathrm{CO}_{2}\) at \(25^{\circ} \mathrm{C}\) is \(3.3 \times 10^{-2} \mathrm{~mol} / \mathrm{L} \cdot \mathrm{atm} ; P_{\mathrm{CO}_{2}}\) in air is \(4 \times 10^{-4}\) atm )?

Gold occurs in seawater at an average concentration of \(1.1 \times 10^{-2}\) ppb. How many liters of seawater must be processed to recover 1 troy ounce of gold, assuming \(81.5 \%\) efficiency \((d\) of seawater \(=1.025 \mathrm{~g} / \mathrm{mL} ; 1\) troy ounce \(=31.1 \mathrm{~g}) ?\)
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