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  • Confused! kindly explain, 36g of glucose, C6H12O6 ,isdissolved in1kg of water in a saucepan. temperature at which water will boil at 1.013 bar? Kb for water is 0.52kg/mole

36g of glucose, C6H12O6 ,is dissolved in 1kg of water in a saucepan. temperature at which water will boil at 1.013 bar? Kb for water is 0.52 kg/mole

  • Option 1)

    373.008K

  • Option 2)

    373.202 K

     

     

  • Option 3)

    373.254K

  • Option 4)

    below 373K

 

Answers (1)

best_answer

As we learned in concept

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

 

 \Delta T_{b}\:=\:K_{b}m

 m=\frac{\frac{36}{180}}{1kg}\:=\:0.2\:mol\:kg^{-1}

\Delta T_{b}\:=\:0.52\times 0.2\:=\:0.104

Water boils at 373.15 K at 1.013 bar pressure

\therefore T = 373.15 + 0.104 = 373.254 K


Option 1)

373.008K

This option is incorrect.

Option 2)

373.202 K

 

 

This option is incorrect.

Option 3)

373.254K

This option is correct.

Option 4)

below 373K

This option is incorrect.

Posted by

Aadil

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