The amount of ice in grams that will separate out on cooling a solution containing 50 g of a non volatile solute - ethylene glycol in 200 g of water to $- 9^{\circ} C$ is:
( $K_{f}$ for water = 1.86 $K k g m o l^{- 1}$ )

48.20 g

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33.33 g

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37.19 g

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4.27 g

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Solution

The correct option is B 33.33 g
$△ T_{f} = \frac{K_{f} \times w_{1} \times 1000}{M_{1} \times w_{2}} \Rightarrow 9 = \frac{1000 \times 1.86 \times 50}{62 \times w_{2}} \Rightarrow w_{2} = 166.67$ g
$∴$ weight of ice separated $= 200 - 166.67 = 33.33$ g

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Similar questions

50 g

200 g

- {9.3}^{\circ} C (K_{f}

= 1.86 {Kkgmol}^{- 1})

50 g

200 g

- {9.3}^{\circ} C (K_{f}

= 1.86 {Kkgmol}^{- 1})

The amount of ice that will separate out on cooling a solution containing 50 gram of Ethylene glycol in 200 gram water to -9.3⁰C

(K_f=1.86K/molal)

- {9.3}^{\circ} C :

[K_{f} f o r w a t e r = 1.86 K m o l^{- 1} k g]

- 9.3^{0} C

K_{f}

K m o l^{- 1} k g

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