A solution containing 62 g ethylene glycol in 250 g water is cooled to -10°C. If Kf for water is 1.86 K kg mol-1, the amount of water (in g) separated as ice is :

Question

A solution containing 62 g ethylene glycol in 250 g water is cooled to -10°C. If Kf for water is 1.86 K kg mol-1, the amount of water (in g) separated as ice is : (A) 32 (B) 48 (C) 16 (D) 64

Tardigrade Admin · Accepted Answer

Δ Tf = kf. m 10 = 1.86 × (62 / 62/Wkg) W = 0.186 kg Δ W = (250 - 186) = 64 gm

Q. A solution containing $62 g$ ethylene glycol in $250 g$ water is cooled to $- 1 0^{\circ} C$ . If $K_{f}$ for water is $1.86 K k g m o l^{- 1}$ , the amount of water (in g) separated as ice is :

32

48

16

64

$Δ T_{f} = k_{f} . m$
$10 = 1.86 \times \frac{62/62}{W _{k g}}$
$W = 0.186 k g$
$Δ W = (250 - 186) = 64 g m$

Q. A solution containing 62g ethylene glycol in 250g water is cooled to −10∘C. If Kf​ for water is 1.86Kkgmol−1, the amount of water (in g) separated as ice is :

32

48

16

64

ΔTf​=kf​.m 10=1.86×Wkg​62/62​ W=0.186kg ΔW=(250−186)=64gm

Q. A solution containing $62 g$ ethylene glycol in $250 g$ water is cooled to $- 1 0^{\circ} C$ . If $K_{f}$ for water is $1.86 K k g m o l^{- 1}$ , the amount of water (in g) separated as ice is :

$Δ T_{f} = k_{f} . m$
$10 = 1.86 \times \frac{62/62}{W _{k g}}$
$W = 0.186 k g$
$Δ W = (250 - 186) = 64 g m$