Chapter 6, Problem 6.111P

6-111 As noted in Section 6-8C, the amount of external pressure that must be applied to a more concentrated solution to stop the passage of solvent molecules across a semipermeable membrane is known as the osmotic pressure The osmotic pressure obeys a law similar in form to the ideal gas law (discussed in Section 5-4), where Substituting for pressure and solving for osmotic pressures gives the following equation: RT MRT, where M is the concentration or molarity of the solution.

(a) Determine the osmotic pressure at 25°C of a 0.0020 M sucrose (C₁₂H₂₂O₁₁) solution.

(b) Seawater contains 3.4 g of salts for every liter of solution. Assuming the solute consists entirely of NaCl (and complete dissociation of the NaCI salt), calculate the osmotic pressure of seawater at 25°C.

(c) The average osmotic pressure of blood is 7.7 atm at 25°C. What concentration of glucose (C₆H₁₂O₆) will be isotonic with blood?

(d) Lysozyme is an enzyme that breaks bacterial cell walls. A solution containing 0.150 g of this enzyme in 210. mL of solution has an osmotic pressure of 0.953 torr at 25°C. What is the molar mass of lysozyme?

(e) The osmotic pressure of an aqueous solution of a certain protein was measured in order to determine the protein's molar mass. The solution contained 3.50 mg of protein dissolved in sufficient water to form 5.00 mL of solution. The osmotic pressure of the solution at 25°C was found to be 1.54 torr. Calculate the molar mass of the protein.

Interpretation Introduction

(a)

Interpretation:

The osmotic pressure of given sucrose solution should be calculated.

Concept Introduction:

Isotonic solutions are solutions in which both the solutions contain same osmolarity. Osmolarity is a term used for multiplication of molarity of the solution with numbers of each particles of the solute. It depends on one gram of solute present in 1000 grams of solution. Osmotic pressure is the pressure which is applied externally on more concentrated solution that stops the movement of solute form semipermeable membrane. Osmotic pressure follows the ideal gas law and following is the equation for it:

$\pi =MRT$

Where,

• Π= osmotic pressure
• M= molarity of compound
• R= the gas constant
• T= temperature in Kelvin

Answer to Problem 6.111P

The sucrose solution contains 0.049atm at $25{}^{°}\text{C}$.

Explanation of Solution

The data given is as follow.

Temperature = $25{}^{°}\text{C}$ =298 K.

Molarity =0.0020M.

From above mentioned equation for osmotic pressure,

$\begin{array}{l}\pi =MRT\\ \pi =0.0020\times 0.0821\times 298\\ \pi =0.0489\approx 0.049\text{ atm}\end{array}$

The osmotic pressure is 0.049atm for given sucrose solution.

Interpretation Introduction

(b)

Interpretation:

The osmotic pressure of seawater at $25{}^{°}\text{C}$ should be calculated.

Concept Introduction:

Isotonic solutions are solutionsin which both the solutions contain same osmolarity. Osmolarity is a term used for multiplication of molarity of the solution with numbers of each particles of the solute. It depends on one gram of solute present in 1000 grams of solution. Osmotic pressure is the pressure which is applied externally on more concentrated solution that stops the movement of solute form semipermeable membrane. Osmotic pressure follows the ideal gas law and following is the equation for it:

$\pi =MRT$

Where,

• Π= osmotic pressure
• M= molarity of compound
• R= the gas constant
• T= temperature in Kelvin

Answer to Problem 6.111P

The osmotic pressure of given NaCl containing seawater solution is 2.845atm.

Explanation of Solution

The data given is follow,

Temperature = $25{}^{°}\text{C}$ =298 K.

NaCl=3.4g per liter.

First calculating molarity of NaCl.

$3.4\text{ g}\times \frac{1\text{ mol}}{58.44\text{ g}}=0.0581\text{ M}$

For each formula, NaCl dissociate into two ions; ${\text{Na}}^{+}$ and ${\text{Cl}}^{-}$, resulting into osmolarity twice of molarity.

Molarity of NaCl solution= $0.0581\times 2=0.1163\text{ M}$

For osmotic pressure of NaCl solution,

$\begin{array}{l}\pi =MRT\\ \pi =0.1163\times 0.0821\times 298\\ \pi =2.845\text{ atm}\end{array}$

The osmotic pressure of given NaCl containing seawater solution is 2.845atm.

Interpretation Introduction

(c)

Interpretation:

The concentration of glucose should be calculated to make it isotonic with blood.

Concept Introduction:

Isotonic solutions are solutionsin which both the solutions contain same osmolarity. Osmolarity is a term used for multiplication of molarity of the solution with numbers of each particles of the solute. It depends on one gram of solute present in 1000 grams of solution. Osmotic pressure is the pressure which is applied externally on more concentrated solution that stops the movement of solute form semipermeable membrane. Osmotic pressure follows the ideal gas law and following is the equation for it:

$\pi =MRT$

Where,

• Π= osmotic pressure
• M= molarity of compound
• R= the gas constant
• T= temperature in Kelvin

Answer to Problem 6.111P

The solution should contain 0.314M of glucose to become isotonic with blood.

Explanation of Solution

The osmotic pressure of blood is 7.7atm at 298K temperature.

To find out the concentration of glucose that should be isotonic with blood,

$\begin{array}{l}\pi =MRT\\ M=\frac{\pi }{RT}\\ M=\frac{7.7}{0.0821\times 298}\\ M=\frac{7.7}{24.4658}\\ M=0.314\text{ M}\end{array}$

The solution should contain 0.314M of glucose to become isotonic with blood.

Interpretation Introduction

(d)

Interpretation:

The molar mass of lysozymes in solution should be calculated.

Concept Introduction:

Isotonic solutions are solutionsin which both the solutions contain same osmolarity. Osmolarity is a term used for multiplication of molarity of the solution with numbers of each particles of the solute. It depends on one gram of solute present in 1000 grams of solution. Osmotic pressure is the pressure which is applied externally on more concentrated solution that stops the movement of solute form semipermeable membrane. Osmotic pressure follows the ideal gas law and following is the equation for it:

$\pi =MRT$

Where,

• Π= osmotic pressure
• M= molarity of compound
• R= the gas constant
• T= temperature in kelvin

Answer to Problem 6.111P

The molar mass of lysozymes is $1.39\times {10}^4\text{ g/mol}$ in solution.

Explanation of Solution

Osmotic pressure =0.953torr= 0.00125atm (1 torr =0.001315atm).

Temperature=298K.

Calculating the moles of lysozyme in solution,

$\begin{array}{l}\pi =\frac{n}{V}RT\\ n=\frac{\pi \times V}{RT}\\ n=\frac{0.00125\times 0.210}{0.0821\times 298}\\ n=\frac{2.625\times {10}^{-4}}{24.4658}\\ n=1.072\times {10}^{-5}\text{ moles}\end{array}$

Now calculating molar mass for lysozyme,

$\begin{array}{c}M=\frac{\text{mass}}{\text{number of moles}}\\ =\frac{0.150}{1.072\times {10}^{-5}}\\ =1.39\times {10}^4\text{ g/mol}\end{array}$.

Interpretation Introduction

(e)

Interpretation:

The molar mass of protein in solution should be calculated.

Concept Introduction:

Isotonic solutions are solutionsin which both the solutions contain same osmolarity. Osmolarity is a term used for multiplication of molarity of the solution with numbers of each particles of the solute. It depends on one gram of solute present in 1000 grams of solution. Osmotic pressure is the pressure which is applied externally on more concentrated solution that stops the movement of solute form semipermeable membrane. Osmotic pressure follows the ideal gas law and following is the equation for it:

$\pi =MRT$

Where,

• Π= osmotic pressure
• M= molarity of compound
• R= the gas constant
• T= temperature in Kelvin

Answer to Problem 6.111P

The molar mass of protein is $8.45\times {10}^3\text{ g/mol}$ in solution.

Explanation of Solution

Osmotic pressure =1.54torr=0.002026atm (1 torr =0.001315atm).

Temperature=298K.

Calculating the moles of protein in solution,

$\begin{array}{l}\pi =\frac{n}{V}RT\\ n=\frac{\pi \times V}{RT}\\ n=\frac{0.002026\times 0.005}{0.0821\times 298}\\ n=\frac{1.031\times {10}^{-5}}{24.4658}\\ n=4.14\times {10}^{-7}\text{ mol}\end{array}$

Now calculating molar mass for lysozyme,

$\begin{array}{c}\text{Molar mass}=\frac{\text{Mass}}{\text{Number of moles}}\\ =\frac{3.5\times {10}^{-3}\text{ g}}{4.14\times {10}^{-7}\text{ mol}}\\ =8.45\times {10}^3\text{ g/mol}\end{array}$.