If $1.71 g$ of sugar (molar mass $= 342$ ) are dissolved in $500 m l$ of an aqueous solution at $300 K$ , what will be its osmotic pressure?

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Solution

Given: Wt. of solute $(w) = 1.71 g$
Molar mass of solute $(m) = 342$
Volume of solution $(v) = 500 m l = 0.5 l t$
Temperature $(T) = 300 K$
Gas constant $(R) = 0.0821 l t a t m K^{- 1}$ ${m o l}^{- 1}$
Osmotic pressure $(π) = \frac{w R T}{m V} = \frac{1.71 \times 0.0821 \times 300}{342 \times 0.5} = 0.25 a t m$
Thus, Osmotic pressure of the solution is $0.25 a t m .$

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If 1.71g of sugar (molar mass=342) are dissolved in 500ml of an aqueous solution at 300K, what will be its osmotic pressure?

Given: Wt. of solute (w)=1.71gMolar mass of solute (m)=342Volume of solution (v)=500ml=0.5ltTemperature (T)=300KGas constant (R)=0.0821ltatmK−1 mol−1Osmotic pressure (π)=wRTmV=1.71×0.0821×300342×0.5=0.25atmThus, Osmotic pressure of the solution is 0.25atm.

If $1.71 g$ of sugar (molar mass $= 342$ ) are dissolved in $500 m l$ of an aqueous solution at $300 K$ , what will be its osmotic pressure?