A solution of 6.2 g ethylene glycol in 55g $H_{2} O$ is cooled to $- {3.72}^{o} C$ . The ice separated from solution is (in g): $(K_{f} H_{2} O = 1.86 K m o l a l i t y^{- 1})$

Open in App

Solution

The molar mass of ethylene glycol is 62 g/mol.
6.2 g of ethylene glycol corresponds to $\frac{6.2 g}{62 g / m o l} = 0.1 m o l$ .
Th relationship between the depression in the freezing point and the molality of the solution is $Δ T_{f} = K_{f} \times m$ .
Hence, $m = \frac{Δ T_{f}}{K_{f}} = \frac{3.72 K}{1.86 K / m} = 2 m$ .
Now 1 m solution contains 1 mole of solute in 1 kg of solvent.
Hence, the mass of water present in 2 molal solution containing 0.1 mole of ethylene glycol will be $\frac{0.1}{2} \times 1000 = 50 g$
Hence, the amount of ice separated will be $55 g - 50 g = 5 g$

Suggest Corrections

Similar questions

50 g

200 g

- {9.3}^{\circ} C

(K_{f}^{'} = 1.86 K m o l a l i t y^{- 1})

[C_{2} H_{4} {(O H)}_{2}]

10^{o} C

K_{f} (H_{2} O)

1.86

- {9.3}^{o} C

[K_{f} = 1.86 K m o l^{- 1}]

- {9.3}^{o} C

[K_{f}

H_{2} O = 1.86

^{- 1}]

50 c m^{3}

- 34^{o} C

(K_{f}

H_{2} O = 1.86 K m o l a l i t y^{- 1})

Join BYJU'S Learning Program