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 :

Options

A. 0.8
B. 0.6
C. 0.65
D. 0.75

Answer

A

Explanation

`DeltaT_(f)=0*3^(@)C " " M_(B)=94"g mol"^(-1)`
`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%`