The osmotic pressure of a sucrose solution containing 40 g of solute (molecular mass 246) per litre at $27^{\circ} C$ is $(R = 0.0822 a t m L m o l^{- 1})$

3.0 atm

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4.0 atm

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2.0 atm

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1.0 atm

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Solution

The correct option is B 4.0 atm
$π = C R T = \frac{(\frac{W_{2}}{M w_{2}}) R T}{V}$
Given $W_{2} = 40 g$
$M w_{2} = 246$
$T = 27^{\circ} C = 300 K$
V = 1 L
Substituting all the values, we get
$π = \frac{40}{246} \times 0.082 \times 300 = 4 a t m$

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Similar questions

Q. The osmotic pressure of a solution containing 40 g of solute (molecular mass 246) per litre at

27^{o} C

is:

[R = 0.0822 a t m L {m o l}^{- 1}]

Q. The Osmotic pressure of solution containing

4.0 g

of solute

(m o l a r m a s s 246)

per liter at

27^{o} C

is :

(R = 0.082 L a t m k^{- 1} m o l^{- 1})

A solution of sucrose (molar mass = 342 g/mol) is prepared by dissolving 68.4 g of it per litre of the solution. What is its osmotic pressure (R = 0.082 lit. atm.k^-1mol^-1) at 273 K?

Q. The osmotic pressure at

17^{\circ} C

of an aqeuous solution containing

1.75 g

of sucrose per

150 m L

solution is

Q. The solution containing

6 g

of an organic compound per litre of the solution shows an osmotic pressure of

1.2 a t m

27^{\circ} C

. Calculate the molecular mass of the compound

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The osmotic pressure of a sucrose solution containing 40 g of solute (molecular mass 246) per litre at 27∘C is (R=0.0822atmLmol−1)

The correct option is B 4.0 atm π=CRT=(W2Mw2)RTV Given W2=40g Mw2=246 T=27∘C=300K V = 1 L Substituting all the values, we get π=40246×0.082×300=4atm

The osmotic pressure of a sucrose solution containing 40 g of solute (molecular mass 246) per litre at $27^{\circ} C$ is $(R = 0.0822 a t m L m o l^{- 1})$

The correct option is B 4.0 atm
$π = C R T = \frac{(\frac{W_{2}}{M w_{2}}) R T}{V}$
Given $W_{2} = 40 g$
$M w_{2} = 246$
$T = 27^{\circ} C = 300 K$
V = 1 L
Substituting all the values, we get
$π = \frac{40}{246} \times 0.082 \times 300 = 4 a t m$