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

`DeltaT_(f)=ixxK_(f)xxm`
`0.3=ixx1.86xx(1.25xx1000)/(50xx94)=0.6064`
`" Degree of association, " alpha=(i-1)/(1/n-1)=(0.6064-1)/(1/2-1)`
`=78.72%~~80%`