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  • Can someone help me with this, At 100°C the vapour pressure of a solution of 6.5 g of a solute in 100 g water is 732mm. If Kb = 0.52, the boiling point of this solution will be:

At 100°C the vapour pressure of a solution of 6.5 g of a solute in 100 g water is 732mm. If Kb = 0.52, the boiling point of this solution will be:

 

  • Option 1)

    101°C

  • Option 2)

    100°C

  • Option 3)

    102°C

  • Option 4)

    103°C    

 

Answers (1)

 

Mathematical Expression -

\Delta T_{b}= K_{b}\: m

Unis of K_{b}=\frac{K-K_{g}}{mole}

\Delta T_{b}= Elevation\: in \: boiling\: point
 

- wherein

K_{b}= Boiling \: point \: elevation \: constant

m= molality

 

 Let's find the molar mass of the solute first By Roalt's law of partial pressure

\Rightarrow \frac{760-732}{732}=\frac{6.5 \times 18}{M \times 100}

\Delta M = 30.6

Then we know that \Delta T_{b}= 0.52 \times \frac{6.5}{30.6} \times \frac{1000}{100}=1.1

\therefore boiling \ point = 100 +1.1 \simeq o1^{\circ}C

 


Option 1)

101°C

correct option

Option 2)

100°C

incorrect option

Option 3)

102°C

incorrect option

Option 4)

103°C    

correct option

Posted by

Sabhrant Ambastha

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