Question

Suppose 0.125 g of a protein is dissolved in $\mathbf{10}\mathbf{.}\mathbf{0}\mathbf{}{\mathbf{\text{cm}}}^{\mathbf{3}}$of ethyl alcohol$\left({\text{C}}_2{\text{H}}_5\text{OH}\right)$ , whose density at ${\mathbf{20}}^{\mathbf{0}}\mathit{C}$ is$\mathbf{0}\mathbf{.}\mathbf{789}\mathbf{}\mathbf{\text{g}}\mathbf{}{\mathbf{\text{cm}}}^{\mathbf{-}\mathbf{3}}$ . The solution rises to a height of26.3 cmin an osmometer (an apparatus for measuring osmotic pressure). What is the approximate molar mass of the protein?

Short answer

The molar mass of the protein is $1.50\times {10}^4{\text{gmol}}^{-1}$

Step-by-step solution

Given data

0.125 g of a protein is dissolved in $10.0{\text{cm}}^3$ of ethyl alcohol $\left({\text{C}}_2{\text{H}}_5\text{OH}\right)$, the density of the protein at ${20}^{0}C$ is $0.789\text{g}{\text{cm}}^{-3}$. The solution rises to a height of 26.3 cm .

Concept of molar mass

The mass of one mole of a sample is its molar mass. The equation shown below can be used to determine the molar mass.

Molar mass (M) = Mass of the substance (g) /Solute concentration (mol) .. (1)

Calculate the  π value

Calculate the$\pi$value to measure the solute concentration. The formula to calculate$\pi$value is;

$\pi =\rho gh$ ....... (2)

$$\begin{gathered} \rho (\text{density of the solution)}=1.00{\text{gcm}}^{-3} \\ g=9.807{\text{ms}}^{-2} \\ h=26.3\text{cm}=0.263 \end{gathered}$$

Substitute these values in equation (2)

$$\begin{gathered} \pi =0.789{\text{gcm}}^{-3}\times {10}^{-3}{\text{kgg}}^{-1}\times {10}^6{\text{cm}}^3{\text{m}}^{-3}\times 9.807{\text{ms}}^{-2}\times 0.263\text{m} \\ =2.04\times {10}^3{\text{kgm}}^{-1}{\text{s}}^{-2} \\ =2.04\times {10}^3\text{Pa} \end{gathered}$$

$$\begin{gathered} \pi =\frac{2.04\times {10}^3\text{Pa}}{1.013\times {10}^5{\text{Paatm}}^{-1}} \\ =0.020\text{atm} \end{gathered}$$

Calculate the solute concentration (C )

The concentration in moles per litre is

$c=\frac{\pi }{RT}$

$$\begin{gathered} \pi =0.020\text{atm} \\ T={20}^0\text{C}+273.15=293.15\text{K} \\ R=0.08206{\text{atmLmol}}^{-1}{\text{K}}^{-1} \end{gathered}$$ ......... (3)

Substitute these values in equation (3)

$$\begin{gathered} c=\frac{0.020\text{atm}}{0.08206\text{atmL}mo{l}^{-1}{\text{K}}^{-1}\times 293.15\text{K}} \\ =8.3\times {10}^{-4}{\text{molL}}^{-1} \end{gathered}$$

Calculate the molar mass

Mass of the substance =12.5 g .

Solute concentration$=8.3\times {10}^{-4}\text{mol}$.

Substitute these values in equation (1) to calculate molar mass

$$\begin{gathered} M=\frac{12.5\text{g}}{8.3\times {10}^{-4}\text{mol}} \\ =1.50\times {10}^4{\text{gmol}}^{-1} \end{gathered}$$