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Solutions question

2019 · 9 Jan · Shift 2 · Q13
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Solutions question

2019 · 9 Jan · Shift 2 · Q13

JEE MainChemistrySolutionsMCQ+4 / −1
A solution containing 62 g ethylene glycol in 250 g water is cooled to −-− 10oC. If Kf for water is 1.86 K kg mol −-− 1 , the amount of water (in g) separated as ice is :
  1. A
    48
  2. B
    32
  3. C
    64
  4. D
    16
View written solutionFree

Correct answer: C

  1. Use freezing point depression relation

For a non-volatile solute: ΔTf=Kfm\Delta T_f = K_f mΔTf​=Kf​m where m=moles of solutekg of solvent remaining in liquid phasem = \frac{\text{moles of solute}}{\text{kg of solvent remaining in liquid phase}}m=kg of solvent remaining in liquid phasemoles of solute​

Since the solution is cooled to −10∘-10^\circ−10∘C, the depression in freezing point is ΔTf=10 K\Delta T_f = 10\,\text{K}ΔTf​=10K

So, m=101.86m = \frac{10}{1.86}m=1.8610​

  1. Calculate moles of ethylene glycol

Ethylene glycol is C2H6O2\mathrm{C_2H_6O_2}C2​H6​O2​. Its molar mass is 2(12)+6(1)+2(16)=24+6+32=62 g mol−12(12) + 6(1) + 2(16) = 24 + 6 + 32 = 62\,\text{g mol}^{-1}2(12)+6(1)+2(16)=24+6+32=62g mol−1

Given mass of ethylene glycol = 626262 g, so moles are n=6262=1 moln = \frac{62}{62} = 1\,\text{mol}n=6262​=1mol

  1. Find mass of water remaining in solution at −10∘-10^\circ−10∘C

Let mass of water remaining unfrozen be WWW g. Then m=1W/1000=1000Wm = \frac{1}{W/1000} = \frac{1000}{W}m=W/10001​=W1000​

Using 10=1.86×1000W10 = 1.86\times \frac{1000}{W}10=1.86×W1000​

So, W=1.86×100010=186 gW = \frac{1.86\times 1000}{10} = 186\,\text{g}W=101.86×1000​=186g

Thus, only 186186186 g water remains in solution.

  1. Calculate ice separated

Initially water = 250250250 g

Water frozen out as ice: 250−186=64 g250 - 186 = 64\,\text{g}250−186=64g

  1. Match with options

64 g\boxed{64\,\text{g}}64g​

So the correct option is C.

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