Question

45 g of ethylene glycol $(C_2H_6O_2)$ is mixed with 600 g of water. The freezing point of the solution is ($K_f$ 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

Answer 1

Answer: (B)

270.95 K

Answer 2

Depression in freezing point is related to the molality, therefore, the molality of the solution with respect to ethylene glycol $= \frac{moles \,of\, ethylene \,glycol}{mass \,of \,water \,in \,kilogram}$ Moles of ethylene glycol $= \frac{45 g }{62\, g \,mol^{-1}} = 0.73 \,mol$ Mass of water in $kg =\frac{600\, kg}{1000 \,kg} = 0.6\, kg$ Hence, molality of ethylene glycol $= \frac{0.73\, mol}{0.60\, kg} = 1.2\, mol\, kg^{-1}$ Therefore, freezing point depression, $?T_{f} = 1.86\, K\, kg\,mol^{-1} ? 1.2 \,mol \,kg^{-1} = 2.2\, K$ Freezing point of the aqueous solution $= 273.15\, K - 2.2\, K = 270.95 \,K$