$x$ g of a non-electrolytic compound (molar mass $= 200$ ) is dissolved in $1.0$ litre of $0.05$ M $N a C l$ solution. The osmotic pressure of this solution is found to be $4.92$ atm at $27^{o}$ C. Calculate the value of $x$ . Assume complete dissociation of NaCl and ideal behaviour of his solution.

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Solution

Molar mass of non electrolute = $M = 200 g$
Number of $m o l e = \frac{x}{200}$
total number of mole of $N a^{+}$ and $C l^{-}$
$2 \times 1 l i t r e \times 0.05 M$
$= 0.1 m o l e$
$C = (\frac{x}{200} + 0.1) / 1 l i t r e$
$π = C R T$
$4.92 = (\frac{x}{200} + 0.1) \times 0.0821 \times 300$
$\frac{4.92}{300 \times 0.0821} = \frac{x}{200} + 0.1$
$0.2 = \frac{x}{200} + 0.1$
$\frac{x}{200} = 0.1$
$x = 20 g$

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x g of a non-electrolytic compound (molar mass =200) is dissolved in 1.0 litre of 0.05M NaCl solution. The osmotic pressure of this solution is found to be 4.92atm at 27oC. Calculate the value of x. Assume complete dissociation of NaCl and ideal behaviour of his solution.

Molar mass of non electrolute = M=200gNumber of mole=x200total number of mole of Na+ and Cl−2×1litre×0.05M=0.1moleC=(x200+0.1)/1litreπ=CRT4.92=(x200+0.1)×0.0821×3004.92300×0.0821=x200+0.10.2=x200+0.1x200=0.1x=20g