A solution of x moles of sucrose in 100 grams of water freezes at -0.2°C As ice separates the freezing point goes down to 0.25°C How many grams of ice would have separated?

Question

A solution of x moles of sucrose in 100 grams of water freezes at -0.2°C As ice separates the freezing point goes down to 0.25°C How many grams of ice would have separated? (A) 18 gram (B) 20 gram (C) 25 gram (D) 23 gram

Tardigrade Admin · Accepted Answer

(i)Δ Tf =m × Kf 0.2=(X×1000/100) ×1.86 X=(0.2/10×1.86) after freezing Δ Tf =m× Kf Δ Tf =(X×1000/(100-y))×1.86 Δ T f =0.25 On solving, Amount of ice y = 20 g ice

Q. A solution of x moles of sucrose in 100 grams of water freezes at - $0. 2^{\circ} C$ As ice separates the freezing point goes down to $0.2 5^{\circ} C$ How many grams of ice would have separated?

18 gram

20 gram

25 gram

23 gram

$(i) Δ T_{f} = m \times K_{f}$
$0.2 = \frac{X \times 1000}{100} \times 1.86$
$X = \frac{0.2}{10 \times 1.86}$
after freezing
$Δ T_{f} = m \times K_{f}$
$Δ T_{f} = \frac{X \times 1000}{( 100 - y )} \times 1.86$
$Δ T_{f} = 0.25$
On solving, Amount of ice y = 20 g ice

Q. A solution of x moles of sucrose in 100 grams of water freezes at -0.2∘C As ice separates the freezing point goes down to 0.25∘C How many grams of ice would have separated?