What should be the freezing point of aqueous solution containing 17 g of C2H5OH in 1000 g of water (water kf=1.86°kg mol-1 ) ?

Question

What should be the freezing point of aqueous solution containing 17 g of C2H5OH in 1000 g of water (water kf=1.86°kg mol-1 ) ? (A)  -0.69°C  (B)  0.34°C  (C)  0.0°C  (D)  0.24°C

Tardigrade Admin · Accepted Answer

Depression in freezing point (Δ Tf) =(1000 × kf × w/m × W) where, kf is molar depression constant, w and W are the weights of solute and solvent, m is the molecular weight of solute. Δ Tf =(1000 × 1.86 × 17/46 × 1000) =0.69° C Freezing point of solution, Tf=0-0.69=-0.69° C

Q. What should be the freezing point of aqueous solution containing $17 g$ of $C_{2} H_{5} O H$ in $1000 g$ of water (water $k_{f} = 1.8 6^{\circ} k g m o l^{- 1}$ ) ?

$- 0.6 9^{\circ} C$

$0.3 4^{\circ} C$

$0. 0^{\circ} C$

$0.2 4^{\circ} C$

Q. What should be the freezing point of aqueous solution containing 17g of C2​H5​OH in 1000g of water (water kf​=1.86∘kgmol−1 ) ?

−0.69∘C

0.34∘C

0.0∘C

0.24∘C

Depression in freezing point (ΔTf​) =m×W1000×kf​×w​ where, kf​ is molar depression constant, w and W are the weights of solute and solvent, m is the molecular weight of solute. ΔTf​=46×10001000×1.86×17​ =0.69∘C Freezing point of solution, Tf​=0−0.69=−0.69∘C

Q. What should be the freezing point of aqueous solution containing $17 g$ of $C_{2} H_{5} O H$ in $1000 g$ of water (water $k_{f} = 1.8 6^{\circ} k g m o l^{- 1}$ ) ?

$- 0.6 9^{\circ} C$

$0.3 4^{\circ} C$

$0. 0^{\circ} C$

$0.2 4^{\circ} C$