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Chemistry20275 marksNumericalSolutionsTool: Molar mass calculator

Answer the following.

Answer the following: 300 ml of aqueous solution of protein contains 1·85 g of protein, the osmotic…

Answer the following:
(a)[1.5]
$300\text{ ml}$ of aqueous solution of protein contains $1\cdot85\text{ g}$ of protein, the osmotic pressure of the solution at $25^\circ\text{C}$ is found to be $3\cdot05 \times 10^{-3}\text{atm}$. Calculate the molar mass of the protein.
(b)[1.5]
Phenol ($\text{C}_6\text{H}_5\text{OH}$) associates in benzene to form a dimer. A solution of $2\cdot5\text{g}$ of phenol in $120\text{g}$ of benzene lowers its freezing point by $0\cdot85\text{K}$. Calculate the degree of association of phenol. ($k_f$ for benzene $= 5\cdot12\text{ K kg mol}^{-1}$, molecular mass of phenol $= 94\text{ g mol}^{-1}$)
(c)[2]
The molecular weight of potassium chloride and sucrose is determined by the depression of freezing point method. Compared to their theoretical molecular weight, what will be their observed molecular weights when determined by the above method? Justify your answer.

Answer

Answer (a)

Official answer key
$\pi = CRT = \dfrac{w\,R\,T}{M\,V}$, with $V = 0.3\text{ L}$, $T = 298\text{ K}$. $M = \dfrac{1.85\times 0.0821\times 298}{3.05\times 10^{-3}\times 0.3} = 49466\text{ g mol}^{-1}$

Final answer: 49466 g mol^-1

Answer (b)

Official answer key

Formula used: Depression of freezing point: dTf = i*Kf*m with Kf = 5.12 K kg mol^-1, w1 = 120 g, w2 = 2.5 g, dTf = 0.85 K, so M2 = 125.49 g mol^-1

$M_{obs} = \dfrac{1000\times K_f\times w}{\Delta T_f\times W} = \dfrac{1000\times 5.12\times 2.5}{0.85\times 120} = 125.49\text{ g mol}^{-1}$ $i = \dfrac{M_{normal}}{M_{obs}} = \dfrac{94}{125.49} = 0.749$ For dimerisation ($n = 2$): $\alpha = \dfrac{1-i}{1-\frac{1}{n}} = \dfrac{1-0.749}{0.5} = 0.502$, i.e. 50.2% association.

Final answer: 0.502

Answer (c)

Official answer key
The observed molecular weight of KCl will be about half its theoretical value, because KCl is a strong electrolyte and ionises into two particles ($\mathrm{KCl \rightarrow K^+ + Cl^-}$); depression of freezing point is a colligative property that depends on the number of particles. The observed molecular weight of sucrose will be the same as its theoretical value, because sucrose is a non-electrolyte and neither ionises nor associates.