What osmotic pressure would the $1.25$ molal sucrose solution exhibit at $25^{o}$ C? (The density of this solution is $1.34$ g/mL.)

28.70

atm

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30.22

atm

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21.72

atm

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24.55

atm

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Solution

The correct option is D $28.70$ atm
Using osmotic pressure formula ,
$n = i n \frac{R T}{V} = i M R T$
For molarity $M = \frac{1000 m \times d}{1000 + m \times M_{w}}$
putting the values
$= \frac{(1000 \times 1.25 \times 1.34)}{(1000 \times 1.25 \times 342)} = 1.733 M$
Hence,
$n = i M R T$
$n = 28.6 a t m$ $(F o r : i = 1)$

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What osmotic pressure would the 1.25 molal sucrose solution exhibit at 25oC? (The density of this solution is 1.34 g/mL.)

The correct option is D 28.70 atmUsing osmotic pressure formula ,n=inRTV=iMRTFor molarity M=1000m×d1000+m×Mwputting the values =(1000×1.25×1.34)(1000×1.25×342)=1.733MHence,n=iMRTn=28.6atm (For:i=1)

What osmotic pressure would the $1.25$ molal sucrose solution exhibit at $25^{o}$ C? (The density of this solution is $1.34$ g/mL.)