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  • (a) Draw the graph between vapour pressure and temperature and explain the elevation in boiling point of a solvent in solution. (b) Determine the osmotic pressure of a solution prepared by dissolving 25mg of <img alt="K_{2}SO_{4}" src="h

(a) Draw the graph between vapour pressure and temperature and explain the elevation in boiling point of a solvent in solution.

(b) Determine the osmotic pressure of a solution prepared by dissolving 25mg of K_{2}SO_{4} in 2 litres of water at 25^{\circ}C assuming it to be completely dissiciated.(Atomic masses K = 39u, S = 32u, O= 16u)

 

 

 

 
 
 
 
 

Answers (1)

(a) On addition of a solute to a solvent, boiling point of the solution increases. Non-volatile solute when added to a solvent, lowers the vapour pressure and hence boiling point increases.

   

\Delta T_{b} = elevation of boiling point

T_{b}^{\circ} = vapour pressure of pure solvent

T_{b} =  vapour pressure of the solution

Addition of non-volatile solute in a solvent, decreases the number of volatile particles on the surface of the solution. Hence, the particles that vapourises out from the surface decreases and hence decreases the vapour pressure of the solution. The elevation of boiling point is given by the formula.

\Delta T_{b} = T_{b}-T_{b}^{\circ}

(b) Weight of the solute (W_{2})=25mg = 25\times 10^{-3}g

Volume of the elevation (V) = 2L

Molar mass of solute K_{2}SO_{4} =174g/mol

Gas constant (R) = 0.0821 LatmK-1

Temperature (T) = 25^{\circ}C = 25+273 = 298K

Osmotic pressure (\pi)= ?

\pi= \frac{W_{2}RT}{M_{2}V}=\frac{25 \times 10^{-3}\times 0.0821 \times 298}{174\times 2}= 1.76 \times 10 ^{-3} atm

Posted by

Sumit Saini

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