A solution of `1.25of 'P'` in 50g of water lawers freezing point by `0.3^(@)C` . Molar mass of 'P' is `94.K_(f("water"))=1.86"K kg mol"^(-1)*` The degree of association of 'P' in water is

Options

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

Answer

A

Explanation

As P undergoes associaiton :
` 2P hArr P_(2)`
If `alpha` is the degree of association .
`{:(,CH_(3)COOH ,hArr,CH_(3)COO^(-), + H^(+)),(,"Initial moles",1,0,0),(,"After dissociation",1-alpha,alpha,alpha):}`
Total number of moles =` 1 - alpha + alpha//2 = 1 - (alpha)/(2)`
`i= (1-alpha//2)/(1)`
Now , observed molar mass ,
`M_(B) = (1000 xx k_(f) xx w_(B))/(w_(A) xx Delta T_(f))`
` = (1000 xx 1.86 xx 1.25 )/(50 xx 0.3) = 155`
` i = ("Normal molar mass")/("Observed molar mass")`
` = (94)/(155) =0.606`
` 1- (alpha)/(2) = 0.606`
` -(alpha)/(2) = 0.606^(-1)`
` = - 0.394`
`alpha = - 0.394`
`alpha = 0.788 or 78.8%`
or `~~ 80%`