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The boiling point of benzene is 353.23 K. When 1.80 g of a non-volatile solute was dissolved in 90 g of benzene, the boiling point is raised to 354.11 K. Calculate the molar mass of the solute. Kb for benzene is 2.53 K kg mol–1

Jayant Kumar , 11 Years ago
Grade 11
anser 2 Answers
Sakshi

Last Activity: 9 Years ago

One can use the fact for the elevation in boiling point on the addition of a non-volatile solute.
The equation to be used is
\Delta Tb= K_{b} * molality

Jerin A Varghese

Last Activity: 5 Years ago

Tb = 354.11 K
T0= 353.23 K
Kb = 2.53 K
wsolute  = 1.8 g
wsolvent  =90 g
\DeltaT = Tb - T0
          =354.11 – 353.23
         =0.88 K
 
\DeltaT =  Kb x wsolute  ÷ wsolvent  x  Mo
 M=  2.53 K x 1.8 g  90 g x 0.88 K 
       = 0.0575 kg/mol
       = 57.5 g/mol
      = 58 g/mol
 
 
 

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