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A solution of 1.25 g of P in 50 g of water lowers freezing point by 0.3^(@)C. Molar mass of P is 94 and K_(f) (water) = 1.86 "K kg mol"^(-1). The degree of association of P if it forms dimers in water is : |
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Answer» 0.8 `K_(f)=1*86"K kg mol"^(-1)""W_(B)=1*25g` `a=?""W_(A)=50g` In order to calculate `alpha`, .`i`. must be known `DeltaT_(f)=(ixxK_(f)xxW_(B)xx1000)/(W_(A)xxM_(B))` `0*3=(ixx1*86xx1*25xx1000)/(50xx94)` or `i=(0*3xx50xx94)/(1*86xx1*25xx1000)=0*606` It is GIVEN that P undergoes ASSOCIATION to form DIMER. `{:(,2P,HARR,P_(2),),("Initial",1,,0,),("after association",1-alpha,,alpha//2,):}` `:.` Total number of moles `=1-alpha+alpha//2` `=1-alpha/2` `i=("Observed number of moles")/("Normal number of moles")` `i=(1-alpha//2)/(1)=1- alpha/2` `0*606=1- alpha/2` `0*606=(2-alpha)/(2)` `1*212=2*alpha` `alpha=2-1*212=0*788` or `78*8%` `~~80%` |
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