31 g of ethylene glycol `(C_2H_6O_2)` is dissolved in 600g of water. The freezing point depression of the solution is `(K_f` for water is 1.86 `K Kg mol^-1)`

Options

A. 3.10K
B. 4.65K
C. 0.77K
D. 1.55K

Answer

D

Explanation

Answer(4) Freezing point of depression, `Delta T_f = K_f (b) (i`)
I = vant's Hoff factor =1 for ethylene glycol .
b = moles of solute/kg of solvent .
1 mole of ethylene glycol = 62g .
31g of ethylene glycol constitutes 0.5 moles (i.e., 31/62 = 0.5) .
Solvent = 600g = 0.6kg .
`therefore b = 0.5/0.6 = 0.833` moles/kg .
`Delta T_f = 1.86 (0.833)(1) = 1.549K cong 1.55k`.