$x$ g of a non-electrolyte (molar mass 20) are dissolved in 1.0 L of 0.05 M $N a C l$ aqueous solution. The osmotic pressure of this mixture was found to be 4.92 atm at $27^{0}$ C. What is the value of $x$ ?

Assume complete dissociation of $N a C l$ and ideal behaviour of this solution.
(Solution constant = 0.082 L atm mol $^{- 1}$ K $^{- 1}$ ).

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Solution

Given that,
$π = 4.92 a t m, T = 27 + 273 = 300 K,$ $V = 1 L$

$n_{1} =$ moles of non electrolyte $= \frac{x}{20}$

$n_{2} =$ moles of $N a C l = 0.05$

Using,
$π V = n_{1} S T + n_{2} (1 + α) S T$

$π V = [n_{1} + n_{2} (1 + α)] S T$ where $α = 1$ for $N a C l$

$\Rightarrow 4.92 \times 1 = (x / 20 + 0.05 \times 2) \times 0.082 \times 300$

or, $\frac{x}{20} = 0.2 - 0.1 = 0.1$

$\Rightarrow x = 0.1 \times 20 = 2 g$

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Given that,π=4.92 atm,T=27+273=300K, V=1Ln1= moles of non electrolyte =x20n2= moles of NaCl=0.05Using, πV=n1ST+n2(1+α)STπV=[n1+n2(1+α)]ST where α=1 for NaCl⇒4.92×1=(x/20+0.05×2)×0.082×300or, x20=0.2−0.1=0.1⇒x=0.1×20=2g