Chapter 5, Problem 5.11E

(a)

Interpretation Introduction

Interpretation:

The density of the solution to be $0.998g\cdot c{m}^{-3}$ and density of mercury is $13.6g\cdot c{m}^{-1}$, molar mass of protein has to be calculated.

Concept introduction:

Molar mass can be calculated by using following formula,

  $\begin{array}{l}\text{Π=}\text{iRT}\times Molarity\\ \\ \text{Where,}\\ \text{Π}\text{-}\text{Osmatic}\text{pressure}\\ \text{T}\text{-}\text{Temperature}\\ \text{R}\text{-}\text{Gas}\text{constant}\end{array}$

Explanation of Solution

Given,

An aqueous solution of $0.010g$ of a protein in $10\text{ mL}$ of water at ${20}^{\text{ o}}C$ shows a rise of 5.22 cm in the apparatus.

Unit conversion is given below,

    $\begin{array}{l}5.22cm=52.2mm\\ density=0.998g\cdot c{m}^{-1}\\ \\ 52.2mm\times \frac{0.998g\cdot c{m}^{-1}}{13.6g\cdot c{m}^{-1}}=3.83mmHg(or)3.83Torr\end{array}$

Osmotic pressure is calculated as follows,

  $\begin{array}{l}\text{Π=}\text{iRT}\times Molarity\\ i=1,(proteinisnonelectrolyte)\\ T=273+20=293K\\ \text{Π}\text{=}1\times \left(\text{0}\text{.0821}\text{L}\cdot \text{atm}\cdot mo{l}^{-1}\cdot {\text{K}}^{\text{-1}}\right)\times \left(293\text{K}\right)\times \frac{\left(\frac{0.010g}{M_{protein}}\right)}{0.010L}=\frac{3.83Torr}{760torr\cdot at{m}^{-1}}\\ \\ M_{protein}=\frac{1\times \left(\text{0}\text{.0821}\text{L}\cdot \text{atm}\cdot mo{l}^{-1}\cdot {\text{K}}^{\text{-1}}\right)\times \left(293\text{K}\right)\times 0.010g\times 760torr\cdot at{m}^{-1}}{0.010L\times 3.83Torr}\\ \\ M_{protein}=\frac{1\times \left(\text{0}\text{.0821}\text{L}\cdot \text{atm}\cdot mo{l}^{-1}\cdot {\text{K}}^{\text{-1}}\right)\times \left(293\text{K}\right)\times 0.010g\times 760torr\cdot at{m}^{-1}}{0.010L\times 3.83Torr}\\ M_{protein}\text{=}4773.38g\cdot mo{l}^{-1}\\ M_{protein}\text{=}4.8\times {10}^{3}g\cdot mo{l}^{-1}\end{array}$

The molar mass of protein is $4.8\times {10}^{3}g\cdot mo{l}^{-1}$.

(b)

Interpretation Introduction

Interpretation:

Freezing point of the solution has to be calculated.

Concept introduction:

Freezing point can be calculated by using following formula,

  $\begin{array}{l}\Delta {\text{T}}_{\text{f}}\text{=}{\text{ik}}_{\text{f}}m\\ \Delta {\text{T}}_{\text{f}}\text{=}{\text{ik}}_{\text{f}}\left(\frac{\frac{m_{solute}}{M_{solute}}}{kgsolvent}\right)\\ \\ \text{Where,}\\ \Delta {\text{T}}_{\text{f}}\text{-}Freezingpo\mathrm{int}depression\\ {\text{k}}_{\text{f}}\text{-}\text{freezing}\text{point}\text{molar constant}\text{of}\text{solvent}\end{array}$

Explanation of Solution

Freezing point is calculated as follows,

  $\begin{array}{l}\Delta {\text{T}}_{\text{f}}\text{=}{\text{ik}}_{\text{f}}\left(\frac{\frac{m_{solute}}{M_{solute}}}{kgsolvent}\right)\\ \\ \Delta {\text{T}}_{\text{f}}\text{=}1\times 1.86K\cdot kg\cdot mo{l}^{-1}\times \frac{\left(\frac{0.010g}{4.8\times {10}^{3}g\cdot mo{l}^{-1}}\right)}{\left(\frac{10mL\times 1.0g\cdot m{L}^{-1}}{1000g\cdot k{g}^{-1}}\right)}\\ \\ \Delta {\text{T}}_{\text{f}}\text{=}3.875\times {10}^{-4}K(or)3.875\times {10}^{-4}{}^{o}C\end{array}$

The freezing point is = ${0.0}^{o}C-3.875\times {10}^{-4}{}^{o}C=-3.875\times {10}^{-4}{}^{o}C$

(c)

Interpretation Introduction

Interpretation:

For calculating molar mass of the large molecule, the best method with reason has to be given.

Explanation of Solution

In the colligative property, osmatic pressure method is the correct method for the calculating molar mass of the large molecule.

Because, the freezing point change is very small. Therefore, it is not possible to measure accurate value of molar mass of the large molecule.