Question

Ethylene glycol is used as an antifreeze in a cold climate Mass of ethylene glycol which should be added to $4$ kg of water to prevent it from freezing at - $6^{\circ}C$ will be $(K_{f}$ for water $= 1.86 \,K \,kg \,mol^{-1}$ and molar mass of ethylene glycol $= 62 \,g\, mol^{-1})$

• A. $800 \,g$
• B. $204.30 \,g$
• C. $400.00 \,g$
• D. $304.60 \,g$

Answer 1

Answer: (A)

$\Delta\,T_{f}$ = Freezing point of $H_{2}O$ - freezing point of ethylene glycol solution
$=0-(-6^{\circ})=6^{\circ}$
$K_{f}=1.86^{\circ}K\,kg\,mol^{-1}$
$w_{1}$ = Mass of ethylene glycol in grams
$w_{2}$ = Mass of solvent $(H_{2}O)$ in grams = 4000 g
$m_{1}$ = Molar mass of ethylene glycol $=62\,g\,mol^{-1}$
$\Delta T_{f} =\frac{1000K_{f} w_{1}}{62\times4000}$
$w_{1}=800\,g$