Question

A protein has been isolated as sodium salt with their molecular formula $N{a}_{x}P$ (this notation means that $xN{a}^{+}$ ions are associated with a negatively charged protein $P^{x-}$). A solution of this salt was prepared by dissolving $0.25g$ of this sodium salt of protein in $10g$ of water and ebulliscopic analysis revealed that solution boils at temperature $5.93\times {10}^{-3\circ }C$ higher than the normal boiling point of pure water. $K_b$ of water $0.52kgmo{l}^{-1}$.

Also, the elemental analysis revealed that the salt contains $1\%$ sodium metal by weight. Determine the value of x and molecular weight of acidic form of protein $H_{x}P$.

• A. $x=18,45561amu$
• B. $x=20,45563amu$
• C. $x=22,45565amu$
• D. $x=24,45567amu$

Answer 1

The correct option is B $x=20,45563amu$

$\Delta T_f=K_f\cdot m\cdot i$

$i=n=x+1$

$m=$ number of moles of solute/weight of solvent in kg

$=\frac{0.25}{M}\times \frac{1000}{10}$

$\Delta T_f=K_{f}mi$

$5.93\times {10}^{-3}=0.52\times \frac{0.25\times 1000}{M\times 10}\times (x+1)$

$\frac{x+1}{M}=4.56\times {10}^{-4}$

Also,

$1\times \frac{M}{100}=23x$

Solving, $x=20$

formula $=H_{x}P=H_{20}P$

Molecule weight

= 20 * 2300 - 20 * 23 + 23 = 45563amu