Correct Answer - A
`K_(3)[Fe (CN)_(6)] rarr 3K^(+) + [Fe (CN)_(6)]^(3-)`
:. i=4
`Delta T_(f) = iK_(f) m = i xx K_(f) xx W_(2)/M_(2) xx 1/W_(1) xx 1000`
`= 4 xx 1.86 xx 0.1 /329 xx 1/100 xx 1000`
`=0.023 = 2.3 xx 10^(-2) .^(@)C or K`
`:. T_(f) = 0 - 2.3 xx 10^(-2) .^(@)C = -2.3 xx 10^(-2) .^(@)C`