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Calculate the molecular weight of cellulose acetate if its 0.5% (wt./vol) solution in acetone (sp. gr. = 0.9) shows an osmotic rise of 23 mm against pure acetone at 27$^{\large\circ}$C.

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0.5% (wt. / vol) solution means 0.5 g of cellulose acetate is dissolved in 100 ml solution.
Osmotic pressure = 23 mm of pure acetone
$\pi$=2.3cm of pure acetone=$\large\frac{2.3\times 0.9}{13.6}$cm of Hg =0.1522 cm of of Hg
$\pi=\large\frac{0.1522}{76}$ atm =0.002 atm
Let the molecular weight of the cellulose acetate be M
Here, number of moles of cellulose acetate (n) $=\large\frac{0.5}{M}$
Volume = (V) = 100 ml = 0.1 lit
R = 0.082 Lit atm $mol^{–1} K^{–1}$, T = (27 + 273) = 300 K
Osmotic pressure($\pi$)=$\large\frac{n}{V}$$RT$
$\Rightarrow \large\frac{0.5}{M}\times \frac{1}{0.1}$$\times 0.0821\times 300$
$\therefore M=61575$
Hence (A) is the correct answer.
answered Jun 11, 2014 by sreemathi.v

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