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Calculate the boiling point of a solution containing 0.45g of camphor (mol. wt. 152) dissolved in 35.4g of acetone (b.p. 56.3°C); Kb per 100 gm of acetone is 17.2°C. IN THIS QUESTION WE USE THE FORMULA = 100 Kb *w2/w1M2 I want to know how we derive this formula in which 100 is in numerator and not 1000 . I know Kb is per 100 g though I am not able to derive this relation 
Asked by govtsecschoolnayaganv051 | 11 Jun, 2019, 01:38: PM
answered-by-expert Expert Answer
Given:
 
Weight of solute, w = 0.45 gm
 
Molecular weight, M = 152
 
Weight of solvent, W = 35.4 gm
 
Molal elevation constant, Kb = 17.2 per 100 gm
 
We have,
 
increment straight T subscript straight b space equals space straight K subscript straight b cross times straight m space space space space space space space straight m space equals space fraction numerator Weight space of space solute space in space kg over denominator Wt. space of space solvent space in space kg cross times Molar space mass space of space solute space in space kg divided by mol end fraction space space space space equals fraction numerator 0.45 cross times 10 to the power of negative 3 end exponent over denominator 35.4 cross times 10 to the power of negative 3 end exponent cross times 152 cross times 10 to the power of negative 3 end exponent end fraction space space space space space equals space 0.083 space straight m increment straight T subscript straight b space equals space straight K subscript straight b cross times straight m space space space space space space space space equals space 1.72 cross times space 0.083 increment straight T subscript straight b space equals 0.1438 space degree straight C
 
 
ΔTb = B.P of solution(T) - B.P of solvent( T0)
 
B.P of solution(T)  =  T0 + Tb 
 
                           = 56.3 + 0.1438
 
 B.P of solution(T) = 56.44 °C
 
The boiling point of solution is 56.44 °C.
Answered by Varsha | 12 Jun, 2019, 10:28: AM

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