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

Question

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

Tardigrade Admin · Accepted Answer

Δ T f = K f × m =1.86 × (1/0.25) ∴ Δ T f =7.44° C Now, new freezing point is -10° C. ∴ 10=1.86 × (1/ W text water ) ∴ W text water =(1.86/10)=0.186 kg =186 g ∴ mass of water separated as ice =250-186 =64 g

Q. A solution containing $62 g$ ethylene glycol in $250 g$ of 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 separated as ice is______ $g$

$Δ T_{f} = K_{f} \times m = 1.86 \times \frac{1}{0.25}$
$∴ Δ T_{f} = 7.4 4^{\circ} C$
Now, new freezing point is $- 1 0^{\circ} C$ .
$∴ 10 = 1.86 \times \frac{1}{W _{water}}$
$∴ W_{water} = \frac{1.86}{10} = 0.186 k g = 186 g$
$∴$ mass of water separated as ice $= 250 - 186$
$= 64 g$

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

ΔTf​=Kf​×m=1.86×0.251​ ∴ΔTf​=7.44∘C Now, new freezing point is −10∘C. ∴10=1.86×Wwater ​1​ ∴Wwater ​=101.86​=0.186kg=186g ∴ mass of water separated as ice =250−186 =64g

Q. A solution containing $62 g$ ethylene glycol in $250 g$ of 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 separated as ice is______ $g$

$Δ T_{f} = K_{f} \times m = 1.86 \times \frac{1}{0.25}$
$∴ Δ T_{f} = 7.4 4^{\circ} C$
Now, new freezing point is $- 1 0^{\circ} C$ .
$∴ 10 = 1.86 \times \frac{1}{W _{water}}$
$∴ W_{water} = \frac{1.86}{10} = 0.186 k g = 186 g$
$∴$ mass of water separated as ice $= 250 - 186$
$= 64 g$