45 grams of ethylene glycol `(C_(2)H_(6)O_(2))` is mixed with 600 grams of water. What is the depression in freezing point (`K_(f)` for water = 1.86 kg `"mole"^(-1)`) (C = 12, O = 16, H = 1 g/mol).

Options

A. 2.25 K
B. 22.5 K
C. 3.25 K
D. 32.5 K

Answer

A

Explanation

M.F `= C_(2)H_(6)O_(2)`
M.wt `= 2(12)+6(1)+2(16)`
`= 62 g mol^(-1)=M_(2)`
Weight of ethylene glycol = 45 g `= W_(2)`
Weight of solvent (water = 600 g `= W_(1)`
`K_(f)` (water) = 1.86 kg `mol^(-1)`
`Delta T = K_(f)m`
`= (K_(f)xx W_(2)xx1000)/(M_(2)xx W_(1))`
`= (1.86xx 45" gram" xx 1000)/(62 g mol^(-1)xx 600 " gram")`
= 2.25 K