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. `75%`

Answer

A

Explanation

1.25 g of P in 50 go `H_2O implies ` 25 g in 1000 g of `H_2O`
Molality `=(25)/(94)`
`Delta t = i xx K_(p) xx m`
`0.3 = ixx1.86 xx (25)/(94)`
`i=(94 xx 0.3)/(1.86 xx 25) = 0.6064`
` alpha = (i-1)/((1)/(n)-1) = (0.6064 -1)/((1)/(2)-1)=(-0.3936)/(-0.5) = 0.787=78.7% = 80%`