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 solution is $1.52$ bar at the same temperature, what would be its concentration?

11 g L^{- 1}

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22 g L^{- 1}

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36 g L^{- 1}

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42 g L^{- 1}

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Solution

The correct option is A $11 g L^{- 1}$
$4.98 = \frac{36 \times S \times 300}{180 \times 1}$
$S = 0.083 b a r$ $L {m o l}^{- 1}$ $K$ $K^{- 1}$
$π_{2} = \frac{n_{2} S T_{2}}{V_{2}}$
$52 = \frac{n_{2} \times 0.083 \times 300}{V_{2}}$
$\frac{n_{2}}{V_{2}} = 0.061 m o l L^{- 1} = 11 g L^{- 1}$

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QUESTION 2.22

At 300 K, 36 g of glucose present in a litre of its solution has an osmotic pressure of 4.98 bar. If the osmotic pressure of the solution is 1.52 bar at the same temperature, what would be its concentration?

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At 300K, 36g of glucose present per litre in its solution has an osmotic pressure of 4.98 bar. If the osmotic pressure of solution is 1.52 bar at the same temperature, what would be its concentration?

The correct option is A 11gL−14.98=36×S×300180×1S=0.083bar Lmol−1 K K−1π2=n2ST2V252=n2×0.083×300V2n2V2=0.061molL−1=11gL−1

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 solution is $1.52$ bar at the same temperature, what would be its concentration?

The correct option is A $11 g L^{- 1}$
$4.98 = \frac{36 \times S \times 300}{180 \times 1}$
$S = 0.083 b a r$ $L {m o l}^{- 1}$ $K$ $K^{- 1}$
$π_{2} = \frac{n_{2} S T_{2}}{V_{2}}$
$52 = \frac{n_{2} \times 0.083 \times 300}{V_{2}}$
$\frac{n_{2}}{V_{2}} = 0.061 m o l L^{- 1} = 11 g L^{- 1}$