The weight of ethyl alcohol which must be added to 1.0 L of water so that the solution will freeze at 14° F is ( Kf of water = 1.86 K kg mol- 1 )

Question

The weight of ethyl alcohol which must be added to 1.0 L of water so that the solution will freeze at 14° F is ( Kf of water = 1.86 K kg mol- 1 ) (A) 263.11 g (B) 247.31 g (C) 236.11 g (D) 281.01 g

Tardigrade Admin · Accepted Answer

texto textC = (( texto textF - 32) × 5/9) 14° F=-10° C So Δ T=10 Δ T=Kf× (wB × 1000/MB × wA) 10=1.86× (wB × 1000/46 × 1000) wB=247.31 g.

Q. The weight of ethyl alcohol which must be added to 1.0 L of water so that the solution will freeze at $1 4^{\circ} F$ is ( $K_{f}$ of water = 1.86 K kg $m o l^{- 1}$ )

263.11 g

247.31 g

236.11 g

281.01 g

$^{o} C = \frac{( ^{o} F - 32 ) \times 5}{9}$

$1 4^{\circ} F = - 1 0^{\circ} C$

So $Δ T = 10$

$Δ T = K_{f} \times \frac{w _{B} \times 1000}{M _{B} \times w _{A}}$

$10 = 1.86 \times \frac{w _{B} \times 1000}{46 \times 1000}$

$w_{B} = 247.31$ g.

Q. The weight of ethyl alcohol which must be added to 1.0 L of water so that the solution will freeze at 14∘F is ( Kf​ of water = 1.86 K kg mol−1 )