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The molal elevation constant of water is 0.51 Km–1 The boiling point of 0.1 molal aqueous NaCl solution is nearly :Options1.100.05°C2.100.1°C3.100.2°C4.101.0°C

Sayeda Mahfuja Rahman , 6 Years ago
Grade 12th pass
anser 2 Answers
Chhavi Jain

Last Activity: 6 Years ago

Dear student , 
 
Use the formula : 
                                      ΔTb = Kb m
 
                                   Tb –  Tbo  = 0.51 * 0.1 =0.051 
                                    
Since water boils at 373.15 K  , So  Tbo = 373.15 K
          
                       Therefore ,  T- 373.15 = 0.051
                                  
                                           Tb  = 373.201  K (  bp of aq. NaCl sol.)
Chemistry Consultant
Askiitians
 
 

Science

Last Activity: 5 Years ago

Tb-100= i*kb*m
 
 
 
 
 
 
Tb=100+(2*0.51*0.1)
 
 
 
 
 
Tb=100.102°C 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
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