45 g of ethylene glycol (C2H6O2) is mixed with 600 g of water. The freezing point of the solution is (Kf for water is 1.86 K kg mol-1)

Question

45 g of ethylene glycol (C2H6O2) is mixed with 600 g of water. The freezing point of the solution is (Kf for water is 1.86 K kg mol-1) (A) 273.95 K (B) 270.95 K (C) 370.95 K (D) 373.95 K

Tardigrade Admin · Accepted Answer

Depression in freezing point is related to the molality, therefore, the molality of the solution with respect to ethylene glycol

= \frac{m o l es o f e t h y l e n e g l yco l}{ma ss o f w a t er in ki l o g r am}

Moles of ethylene glycol

= \frac{45 g}{62 g m o l ^{- 1}} = 0.73 m o l

Mass of water in

k g = \frac{600 k g}{1000 k g} = 0.6 k g

Hence, molality of ethylene glycol

= \frac{0.73 m o l}{0.60 k g} = 1.2 m o l k g^{- 1}

Therefore, freezing point depression,

Δ T_{f} = 1.86 K k g m o l^{- 1} \times 1.2 m o l k g^{- 1} = 2.2 K

Freezing point of the aqueous solution

= 273.15 K - 2.2 K = 270.95 K

Q. 45 g of ethylene glycol (C2​H6​O2​) is mixed with 600 g of water. The freezing point of the solution is (Kf​ for water is 1.86Kkgmol−1)

273.95 K

270.95 K

370.95 K

373.95 K

Q. 45 g of ethylene glycol $(C_{2} H_{6} O_{2})$ is mixed with 600 g of water. The freezing point of the solution is ( $K_{f}$ for water is $1.86 K k g m o l^{- 1}$ )