Question

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

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

Answer 1

Answer: (B)

Salt of protein undergoes dissociation as follows.
$Na _{20} P \longrightarrow 20 Na ^{+}+ P ^{20-}$
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 $M =4.5 \times 10^{4} \,g \,mol ^{-1}$