At 300 K, 36 g of glucose present per litre in its solution has an osmotic pressure of 4.98 bar. If the osmotic pressure of the solution is 1.58 bar at the same temperature, what would be its concentration?

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Solution

Osmotic pressure $(π) = C R T$

$\frac{π_{1}}{π_{2}} = \frac{C_{1} R T}{C_{2} R T}$ { As temperature is same}
$\frac{π_{1}}{π_{2}} = \frac{C_{1}}{C_{2}}$

$\Rightarrow \frac{4.96 b a r}{1.58 b a r} = \frac{\frac{36}{180 \times 1}}{C_{2}}$

$C_{2} = \frac{36 \times 1.58}{180 \times 4.96} = 0.063 M$

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At 300 K, 36 g of glucose present per litre in its solution has an osmotic pressure of 4.98 bar. If the osmotic pressure of the solution is 1.58 bar at the same temperature, what would be its concentration?

Osmotic pressure(π)=CRTπ1π2=C1RTC2RT { As temperature is same}π1π2=C1C2⇒4.96bar1.58bar=36180×1C2C2=36×1.58180×4.96=0.063M