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Q.
How much ethyl alcohol $(C_{2}H_{5}OH)$ must be added to $1\,L$ of water so that solution will not freeze at $-6^{\circ}F$ ? $(K_{f}$ of $H_{2}O=1.86^{\circ}\,mol^{-1}\,kg)$
Solutions
Solution:
$-6^{\circ}F$ is to be converted into $^{\circ}C$
$\frac{F-32}{180} \frac{C}{100}$
$\therefore C=-27^{\circ}C$
$\therefore \Delta T_{f} =-27^{\circ}C$
$1\,L\,H_{2}O=1000\,g$ water
$\Delta T_{f} =\frac{1000\,K_{f} w_{1}}{m_{1} w_{1}}$
$w_{1}$ (ethyl alcohol) $=\frac{\Delta T_{f} m_{1} w_{2}}{1000k_{f}}=\frac{27\times46\times1000}{1000\times1.86}$
$=\frac{12466}{186}=670\,g$