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  • Please help! The latent heat of vapourisation of water is 9700Cal / mole and if the b.p. is 1000C , ebullioscopic constant of water is

The latent heat of vapourisation of water is 9700Cal / mole and if the b.p. is  1000C , ebullioscopic constant of water is

  • Option 1)

    0.513^{\circ}C

     

     

     

  • Option 2)

    1.026^{\circ}C

  • Option 3)

    10.26^{\circ}C

  • Option 4)

    1.832^{\circ}C

 

Answers (1)

As we learned 

 

Dependence of K_{b} -

K_{b}  depends only on nature of solvent.

K_{b} =\frac{ RT_{b}^{2}\cdot M}{1000\, \Delta H_{vop}}

imp

(Use value of R in accordance to unit of \Delta H_{vap})
 

 

- wherein

T_{b} = Boiling point of pure solvent in gm

    M = Molar mass of solvent in gm

\Delta H_{vap}= Enthalpy of vap

R = Universal gas constant

 

 

K_{b}=\frac{M_{1}RT_{0}^{2}}{1000\Delta H_{v}}=\frac{18\times 1.987\times (373)^{2}}{1000\times 9700}=0.513^{\circ}C

 


Option 1)

0.513^{\circ}C

 

 

 

Option 2)

1.026^{\circ}C

Option 3)

10.26^{\circ}C

Option 4)

1.832^{\circ}C

Posted by

Vakul

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