A solution containing `0.85 ` g of `ZnCl_(2)` in `125.0` g of watcr freezcs at `-0.23^(@)C.` The apparent degree of dissociation of the salt is `(K _(f)` for water = `1.86 k kg mol ^(-1),` atomic mas: `Zn =65.3 and Cl =35.5)`

Options

A. `1.36%`
B. `73.5%`
C. `7.35%`
D. `2.47%`

Answer

B

Explanation

Mol wt. `= (k _(f) xx w xx 1000)/( Delta T _(f) xx W)`
`= (1.86xx0.85 xx1000)/( 0.23xx125)~~55 gm`
where
`w= 0.85g`
`W= 125g`
`Delta T_(f) =0^(@)C-(-23^(@)C)=23^(@)C`
Now, `i= (M_("normal"))/(M_("obscrved")) = (136.3)/(55) =2.47`
{:(Zn Cl_(2), hArr, Zn^(++), +, 2Cl^(-)), (1-alpha,, alpha,, 2 alpha):}`
Van't Hoff factor (i)
`= (1- alpha _(eta) + alpha _(eta) +2 alpha_(eta))/(1) =2.47`
`therefore alpha eta = 0.735 = 73.5%`