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Standard XII

Chemistry

Osmotic Pressure

Osmotic press...

Question

Osmotic pressure of $30 %$ solution of glucose is $1.20$ atm and that of $3.42 %$ solution of cane sugar is $2.5$ atm. The osmotic pressure of the mixture containing equal volumes of the two solutions will be:

2.5

atm

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3.7

atm

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1.85

atm

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1.3

atm

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Solution

The correct option is D $1.85$ atm
$M (2 V) = M_{1} V + M_{2} V$

$M = \frac{M_{1} + M_{2}}{2}$

$M_{1} = \frac{1.2}{S T}$

$M_{2} = \frac{2.5}{S T}$

$M = \frac{3.7}{2 S T} = \frac{1.85}{S T}$

$\Rightarrow M \times S \times T = 1.85 a t m$

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Osmotic pressure of 30% solution of glucose is 1.20 atm and that of 3.42% solution of cane sugar is 2.5atm. The osmotic pressure of the mixture containing equal volumes of the two solutions will be:

The correct option is D 1.85atmM(2V)=M1V+M2VM=M1+M22M1=1.2STM2=2.5STM=3.72ST=1.85ST⇒M×S× T=1.85atm

Osmotic pressure of $30 %$ solution of glucose is $1.20$ atm and that of $3.42 %$ solution of cane sugar is $2.5$ atm. The osmotic pressure of the mixture containing equal volumes of the two solutions will be:

The correct option is D $1.85$ atm
$M (2 V) = M_{1} V + M_{2} V$

$M = \frac{M_{1} + M_{2}}{2}$

$M_{1} = \frac{1.2}{S T}$

$M_{2} = \frac{2.5}{S T}$

$M = \frac{3.7}{2 S T} = \frac{1.85}{S T}$

$\Rightarrow M \times S \times T = 1.85 a t m$