Calculate osmotic pressure of 3-42- w-V solution of cane sugar sucrose at 27-C-Given-Molecular formula of sucrose is C12H22O11

nbsp;Molecular mass of sucrose =(12×12)+(22×1)+(11× 16)=342 g mol^ 1 3.42 g of sucrose is present in 100 mL of solution. Osmotic pressure : π=nV× RT Here, n ...

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Standard XII

Chemistry

Osmotic Pressure

Calculate osm...

Question

Calculate osmotic pressure of $3.42 % w / V$ solution of cane sugar (sucrose) at $27^{\circ} C$ .
Given:
$Molecular formula of sucrose is C_{12} H_{22} O_{11}$

0.16 a t m

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0.08 a t m

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2.46 a t m

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4.96 a t m

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Solution

The correct option is C $2.46 a t m$
$Molecular mass of sucrose = (12 \times 12) + (22 \times 1) + (11 \times 16) = 342 g m o l^{- 1}$
$3.42 g$ of sucrose is present in $100 m L$ of solution.

Osmotic pressure :
$π = \frac{n}{V} \times R T$
Here,
$n = \frac{3.42}{342} = 0.01 V = 100 m L = 0.1 L R = 0.0821 L a t m m o l^{- 1} K^{- 1} T = 27^{\circ} C = 300 K$

$π = \frac{0.01}{0.1} \times 0.0821 \times 300 π = 2.46 a t m$

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What is the osmotic pressure of a solution containing 3.42 g of cane sugar in 1litre solution at 27^oC .Molar mass of cane sugar is 342 g.

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Osmotic Pressure

Standard XII Chemistry

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Calculate osmotic pressure of 3.42% w/V solution of cane sugar (sucrose) at 27∘C. Given: Molecular formula of sucrose is C12H22O11

The correct option is C 2.46 atm Molecular mass of sucrose =(12×12)+(22×1)+(11×16)=342 g mol−1 3.42 g of sucrose is present in 100 mL of solution. Osmotic pressure : π=nV×RT Here, n=3.42342=0.01V=100 mL=0.1 LR=0.0821 L atm mol−1 K−1T=27∘C=300 K π=0.010.1×0.0821×300π=2.46 atm

Calculate osmotic pressure of $3.42 % w / V$ solution of cane sugar (sucrose) at $27^{\circ} C$ .
Given:
$Molecular formula of sucrose is C_{12} H_{22} O_{11}$