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. `K_("(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

Given : `K_(f)=1.86 "K kg mol"^(-1), w=1.25g`,
`W=50g, DeltaT=0.3^(@)C, M=?`
As P undergoes association,
`2P hArr (P)_(2),n=2`
Using equation,
`M=(1000xxK_(f)xx w)/(W xx Delta T)`
`M=(1000xx1.86xx1.25)/(50xx0.3)=155`
Now, `i=("Normal mol. mass")/("Observel mol. mass")=(94)/(155)=0.606`
Then, the degree of association of P is
`alpha=(1-i)/(1-1/n)`
`alpha=(1-0.606)/(1- 1/2)= 0.788` or `78.8% ~80%`