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Standard XII

Chemistry

Characteristics of Equilibrium Constant

A solution is...

Question

A solution is prepared by dissolving $0.6 g$ of urea $(molar mass = 60 g m o t^{- 1}) a n d 1.8 g$ of glucose $(molar mass = 180 g m o l^{- 1} i n 100 m L$ of water at $27^{\circ} C$ . The osmotic pressure of the solution is $(R = 0.08206 L a t m K^{- 1} m o l^{- 1})$

4.92 a t m

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2.46 a t m

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1.64 a t m

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8.2 a t m

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Solution

The correct option is A $4.92 a t m$
Osmotic pressure $(π)$ = CRT

Solute : urea and glucose

$∴ π = (C_{1} + C_{2}) R T$

$= (\frac{0.6}{60 \times 0.1} + \frac{1.8}{180 \times 0.1}) \times 0.08206 \times 300$

$= 0.2 \times 0.0821 \times 300$

$= 4.926 a t m$

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Similar questions

Q. A solution is prepared by dissolving

0.6 g

of urea (molar mass =

60 g m o l^{- 1}

) and

1.8 g

of glucose (molar mass =

180 g m o l^{- 1}

) in

100 m L

of water at

27^{0} C

. The osmotic pressure of the solution is :

(R = 0.08206 L a t m K^{- 1} m o l^{- 1})

Q. A solution of sucrose (Molar mass

= 342 g / m o l)

is prepared by dissolving

68.4 g

of it per litre of solution, what is osmotic pressure

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

273 K

Q. Determine the osmotic pressure of a solution a solution prepared by dissolving

2.5 \times 10^{- 2} g

K_{2} S O_{4}

in 2L of water at

25^{0} C

, assuming that it is completely dissociated.
(

R = 0.0821 L a t m

K^{- 1} m o l^{- 1}

, Molar mass of

K_{2} S O_{4} = 174 g m o l^{- 1}

)

6.0 g

of urea

(Molar mass = 60 g / m o l)

was dissolved in

9.9 m o l

of water. If the vapour pressure of pure water is

P_{A}^{\circ},

the vapour pressure of solution is:

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?

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A solution is prepared by dissolving 0.6 g of urea (molar mass =60 g mot−1) and 1.8 g of glucose (molar mass =180 g mol−1 in 100 mL of water at 27∘C. The osmotic pressure of the solution is (R=0.08206 L atm K−1 mol−1)

The correct option is A 4.92 atmOsmotic pressure (π) = CRT Solute : urea and glucose ∴ π=(C1+C2)RT =(0.660×0.1+1.8180×0.1)×0.08206×300 =0.2×0.0821×300 =4.926 atm

A solution is prepared by dissolving $0.6 g$ of urea $(molar mass = 60 g m o t^{- 1}) a n d 1.8 g$ of glucose $(molar mass = 180 g m o l^{- 1} i n 100 m L$ of water at $27^{\circ} C$ . The osmotic pressure of the solution is $(R = 0.08206 L a t m K^{- 1} m o l^{- 1})$

The correct option is A $4.92 a t m$
Osmotic pressure $(π)$ = CRT

Solute : urea and glucose

$∴ π = (C_{1} + C_{2}) R T$

$= (\frac{0.6}{60 \times 0.1} + \frac{1.8}{180 \times 0.1}) \times 0.08206 \times 300$

$= 0.2 \times 0.0821 \times 300$

$= 4.926 a t m$