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Q.
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 $1.86\, K\, kg\, mol^{-1}$)
Solutions
Solution:
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,
$ΔT_{f} = 1.86\, K\, kg\,mol^{-1} × 1.2 \,mol \,kg^{-1} = 2.2\, K$
Freezing point of the aqueous solution
$= 273.15\, K - 2.2\, K = 270.95 \,K$