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A solution of 1.25of 'P' in 50g of water...

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

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The correct Answer is:
80
As P undergoes association
`2P hArr P_2`
If `alpha` is the degree of association
`{:("Initial",1,0),("After association",1 -alpha, alpha//2):}`
Total number of moles `=1-alpha+alpha//2=1-alpha/2`
`i=(1-alpha//2)/(1)`
Now, observed molar mass,
`M = (1000 xx K_f xx w_B)/(w_A xx DeltaT_f)`
`= (1000 xx 1.86 xx1.25)/(50xx0.3) = 155`
`i=("Normal molar mass")/("Observed molar mass")`
`= 94/155 = 0.606`
`1 - alpha/2 = 0.606`
`=-0.394`
`:. alpha =0.788` or 78.8%
or = 80%.
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