Question

A protein has been isolated as a salt with the formula $N{a}_{20}P$ (this notation means that there are $20N{a}^{+}$ions associated with a negatively charged protein $P^{20-}$ ) The osmotic pressure of a $10.0ml$ solution containing $0.225g$ of the protein is $0.257atm$ at $25^{\circ }C$. The actual molar mass of the protein from these data is approximately.

• A. $2.6\times 10^{5}gmo{l}^{-1}$
• B. $4.5\times 10^{4}gmo{l}^{-1}$
• C. $4.5\times 10^{6}gmo{l}^{-1}$
• D. $2.3\times 10^{3}gmo{l}^{-1}$

Answer 1

Answer: (B)

Salt of protein undergoes dissociation as follows.
$N{a}_{20}P⟶20N{a}^{+}+P^{20-}$
The formula of protein will be $H_{20}P$
Total concentration of ions in the solution
$=$ concentration of cations $+$ conc. of anion
$=20\left[H_{20}P\right]+\left[H_{20}P\right]$
$=(20+1)\left[H_{20}P\right]$
$=21\left[H_{20}P\right]$
$\pi V=nRT$
$\pi =CRT$
$C$ is total concentration of all ions present in the solution
$\left[H_{20}P\right]=\frac{0.225}{M}\times \frac{1000}{10}$
$C=21\times \frac{0.225}{M}\times \frac{1000}{10}$
$0.257=21\times \frac{0.225}{M}\times \frac{1000}{10}\times 0.0821\times 298$
$M=\frac{21\times 0.225\times 0.08821\times 298\times 1000}{0.257\times 10}$
On solving $M=4.5\times 10^{4}gmo{l}^{-1}$