Chapter 13, Problem 149AP

Interpretation Introduction

Interpretation:

The molar mass of protein and actual mass of protein is to be calculated with given mass, volume, pressure, and temperature.

Concept Introduction:

The property that depends on the concentration of solute molecules and not on the identity of solute is termed as colligative property.

The four colligative properties of solution are vapor-pressure lowering, boiling point elevation, freezing point depression, and osmotic pressure.

Osmotic pressure (πp) of a solution is the product of molarity, gas constant, and temperature. It is expressed as follows:

$\pi =M\text{R}T$

Here, $\pi$ is the osmotic pressure, $M$ is the molarity, $\text{R}$ is the gas constant, and $T$ is the temperature.

The molarity of a solution can be expressed as

$M\text{ = }\frac{n}{V}$

Here, $M$ is the molarity, $n$ is the number of moles, and $V$ is the volume in liters.

The molar mass of the protein is calculated as

$M=\frac{m}{n}$

Here, $M$ is the molar mass, $m$ is the given mass, and $n$ is the number of moles.

Answer to Problem 149AP

Solution:

(a)

$2.14\times {10}^3\text{ g/mol}$

(b)

$4.50\times {10}^4\text{ mol}$

Explanation of Solution

a) The molar mass of protein from given data

Calculate the molarity of the solution using the equation given below:

$\text{M}=\frac{\pi }{\text{R}\text{T}}$

Substitute $\text{0}\text{.257 atm}$ for $\pi$, $0.0821\text{ L}\text{.atm}{\text{.mol}}^{\text{-1}}{\text{.K}}^{\text{-1}}$ for $\text{R}$, and $\text{298 K}$ for $T$

$\begin{array}{c}\text{M}=\frac{0.257\cancel{atm}}{(0.0821L\cdot \cancel{\text{atm}}\text{/mol}\cdot \cancel{K})(298\cancel{K})}\\ =0.0105\text{M}\end{array}$

This is the concentration of all the ions. The moles of solute dissolved are calculated as

$M\text{ = }\frac{n}{V}$

Here, $M$ is the molarity, $n$ is the number of moles, and $V$ is the volume in liters.

Rearrange the above equation and substitute the values of molarity and volume:

$\begin{array}{c}n=\left(\frac{0.0105\text{ mol}}{1\cancel{\text{L}}}\times 0.0100\cancel{\text{L}}\right)\\ =1.05\times {10}^{-4}\text{mol}\end{array}$

The molar mass of the protein is calculated as

$M=\frac{m}{n}$

Here, $M$ is the molar mass, $m$ is the given mass and $n$ is the number of moles.

Substitute $0.225\text{ g}$ for $m$ and $1.05\times {10}^{-4}\text{ mol}$ for $n$

$\begin{array}{c}M=\frac{0.225\text{ g}}{1.05\times {10}^{-4}\text{ mol}}\\ =2.14\times {10}^3\text{ g/mol}\end{array}$

Therefore, the molar mass of protein is $2.14\times {10}^3\text{ g/mol}$.

b) The actual molar mass of protein

Protein is a strong electrolyte. The Van’t Hoff factor will be $21$.

Calculate the molarity of the solution using the equation given below:

$\text{M}=\frac{\pi }{i\text{R}\text{T}}$

Substitute $\text{0}\text{.257 atm}$ for $\pi$, $0.0821\text{ L}\text{.atm}{\text{.mol}}^{\text{-1}}{\text{.K}}^{\text{-1}}$ for $\text{R}$, $21$ for $i$, and $\text{298 K}$ for $T$

$\begin{array}{c}\text{M}=\frac{0.257\cancel{atm}}{\left(21\right)(0.0821L\cdot \cancel{\text{atm}}\text{/mol}\cdot \cancel{K})(298\cancel{K})}\\ =5.00\times {10}^{-4}\text{M}\end{array}$

This is the actual concentration of the protein. The moles of solute dissolved are calculated as

$M\text{ = }\frac{n}{V}$

Here, $M$ is the molarity, $n$ is the number of moles, and $V$ is the volume in liters.

Rearrange the above equation and substitute the values of molarity and volume

$\begin{array}{c}n=\left(\frac{5.00\times {10}^{-4}\text{ mol}}{1\cancel{\text{L}}}\times 0.0100\cancel{\text{L}}\right)\\ =5.00\times {10}^{-6}\text{mol}\end{array}$

The actual molar mass is calculated as

$M=\frac{m}{n}$

Here, $M$ is the molar mass, $m$ is the given mass, and $n$ is the number of moles.

Substitute $0.225\text{ g}$ for $m$ and $5.00\times {10}^{-6}\text{ mol}$ for $n$

$\begin{array}{c}M=\frac{0.225\text{ g}}{5.00\times {10}^{-6}\text{ mol}}\\ =4.50\times {10}^4\text{ g/mol}\end{array}$

Therefore, the molar mass of protein is $4.50\times {10}^4\text{ g/mol}$.