How much ethyl alcohol (C2H5OH) must be added to 1 L of water so that solution will not freeze at -6°F ? (Kf of H2O=1.86° mol-1 kg)

Question

How much ethyl alcohol (C2H5OH) must be added to 1 L of water so that solution will not freeze at -6°F ? (Kf of H2O=1.86° mol-1 kg) (A) 122 g (B) 512 g (C) 670 g (D) 495 g

Tardigrade Admin · Accepted Answer

-6°F is to be converted into °C (F-32/180) (C/100) ∴ C=-27°C ∴ Δ Tf =-27°C 1 L H2O=1000 g water Δ Tf =(1000 Kf w1/m1 w1) w1 (ethyl alcohol) =(Δ Tf m1 w2/1000kf)=(27×46×1000/1000×1.86) =(12466/186)=670 g

Q. How much ethyl alcohol $(C_{2} H_{5} O H)$ must be added to $1 L$ of water so that solution will not freeze at $- 6^{\circ} F$ ? $(K_{f}$ of $H_{2} O = 1.8 6^{\circ} m o l^{- 1} k g)$

$122 g$

$512 g$

$670 g$

$495 g$

Q. How much ethyl alcohol (C2​H5​OH) must be added to 1L of water so that solution will not freeze at −6∘F ? (Kf​ of H2​O=1.86∘mol−1kg)

122g

512g

670g

495g

−6∘F is to be converted into ∘C 180F−32​100C​ ∴C=−27∘C ∴ΔTf​=−27∘C 1LH2​O=1000g water ΔTf​=m1​w1​1000Kf​w1​​ w1​ (ethyl alcohol) =1000kf​ΔTf​m1​w2​​=1000×1.8627×46×1000​ =18612466​=670g

Q. How much ethyl alcohol $(C_{2} H_{5} O H)$ must be added to $1 L$ of water so that solution will not freeze at $- 6^{\circ} F$ ? $(K_{f}$ of $H_{2} O = 1.8 6^{\circ} m o l^{- 1} k g)$

$122 g$

$512 g$

$670 g$

$495 g$