When 20 g of naphthoic acid `(C_(11)H_(8)O_(2))` is dissolved in 50 g of benzene (`K_(f) = 1.72 K "kg" mol^(-1)`) , a freezing point depression of 2K is observed . The van't Hoff factor (i) is :

Options

A. `0.5`
B. 1
C. 2
D. 3

Answer

A

Explanation

Molecular weight of napthoic acid, `C_(11)H_(8)O_(2)` is `172 g mol^(-1)`
The theoretical value of depression in freezing point = `k_(f)xx` molality `=1.72xx(20xx1000)/(172xx50)=4K`
`therefore` Van't Hoff factor i=
`("Observed value of colligative property")/("Theoretical value of colligative property")=(2)/(4)=0.5`