Relationship between osmotic pressure at 273 K when 1- glucose -1- 1- urea -2- 81- sucrose -3 are dissolved in 1 litre of water-

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Standard XII

Chemistry

Osmotic Pressure

Relationship ...

Question

Relationship between osmotic pressure at 273 K when $1 %$ glucose ( $π_{1}$ ), $1 %$ urea ( $π_{2}$ ), $81 %$ sucrose ( $π_{3}$ ) are dissolved in 1 litre of water:

π_{1} > π_{2} > π_{3}

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π_{2} > π_{1} > π_{3}

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π_{3} > π_{1} > π_{2}

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π_{1} = π_{2} = π_{3}

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Solution

The correct option is D $π_{2} > π_{1} > π_{3}$
we have,
$π = i C R T$
i=1 for all these non-electrolyte substances.
1% glucose means 1 g of glucose in 100 g of solution,
so, concentration of glucose = $\frac{1}{180}$
so, $π_{1} = \frac{1}{180} R T$

similarly, for 1% urea , $π_{2} = \frac{1}{90} R T$

and for 1% sucrose, $π_{3} = \frac{1}{342} R T$

Hence, we get $π_{2} > π_{1} > π_{3}$

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Relationship between osmotic pressure at 273 K when 1% glucose (π1), 1% urea (π2), 81% sucrose (π3) are dissolved in 1 litre of water:

The correct option is D π2>π1>π3we have,π=iCRTi=1 for all these non-electrolyte substances.1% glucose means 1 g of glucose in 100 g of solution,so, concentration of glucose = 1180so, π1=1180RTsimilarly, for 1% urea , π2=190RTand for 1% sucrose, π3=1342RTHence, we get π2>π1>π3

Relationship between osmotic pressure at 273 K when $1 %$ glucose ( $π_{1}$ ), $1 %$ urea ( $π_{2}$ ), $81 %$ sucrose ( $π_{3}$ ) are dissolved in 1 litre of water: