Question

18 gram of glucose is dissolved in 1 kg of water in a sociopath at what temperature will the water boil at 1.013 bar pressure kb for water is 0.52

Answer 1

Calculations :   

Moles of glucose  ,  C6 H12 O6 

=   Mass of glucose  /   malar mass of glucose

=    1818018180 mol

=  0.1 mol

Number of Kg of solvent  

=  1 kg ........(given)

∴∴molality (m)  of glucose solution 

=   0.1 mol kg-1 

For water , change in boiling point is given by the expression ,

ΔTΔTb    =    Kb x m

where  ,   Kb is Molal boiling point elevation constant is known as  

=  0.52  K  kg mol-1   

and  m  represents molality

=  0.1 mol kg-1   ......calculated as above 

So  , substituting the values in above expression we get ,

ΔTb=0.52kgmol−1X0.1molkg−1ΔTb=0.52kgmol−1X0.1molkg−1

=  0.052 K

Since water boils at 373.15 K at 1.013 bar pressure , therefore the boiling point of solution will be 

=  ( 373.15  +   0.052 ) K

=373.202 K