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Solutions question

2021 · 27 Aug · Shift 2 · Q17
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Solutions question

2021 · 27 Aug · Shift 2 · Q17

JEE MainChemistrySolutionsNumerical+4 / −1
40 g of glucose (Molar mass = 180) is mixed with 200 mL of water. The freezing point of solution is ‾\underline{\hspace{2cm}}​ K. (Nearest integer) [Given : Kf = 1.86 K kg mol −-− 1; Density of water = 1.00 g cm −-− 3; Freezing point of water = 273.15 K]
Numerical answer
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Correct answer: 271

  1. Given data
  • Mass of glucose =40 g= 40\text{ g}=40 g
  • Molar mass of glucose =180 g mol−1= 180\text{ g mol}^{-1}=180 g mol−1
  • Volume of water =200 mL= 200\text{ mL}=200 mL
  • Density of water =1.00 g cm−3= 1.00\text{ g cm}^{-3}=1.00 g cm−3
  • Kf=1.86 K kg mol−1K_f = 1.86\text{ K kg mol}^{-1}Kf​=1.86 K kg mol−1
  • Freezing point of pure water =273.15 K= 273.15\text{ K}=273.15 K
  1. Mass of solvent (water)

Since density of water is 1.00 g mL−11.00\text{ g mL}^{-1}1.00 g mL−1,

Mass of water=200 mL×1.00 g mL−1=200 g=0.200 kg\text{Mass of water} = 200\text{ mL} \times 1.00\text{ g mL}^{-1} = 200\text{ g} = 0.200\text{ kg}Mass of water=200 mL×1.00 g mL−1=200 g=0.200 kg

  1. Moles of glucose

n=40180=29≈0.2222 moln = \frac{40}{180} = \frac{2}{9} \approx 0.2222\text{ mol}n=18040​=92​≈0.2222 mol

  1. Molality of solution

m=moles of solutekg of solvent=0.22220.200=1.111 mol kg−1m = \frac{\text{moles of solute}}{\text{kg of solvent}} = \frac{0.2222}{0.200} = 1.111\text{ mol kg}^{-1}m=kg of solventmoles of solute​=0.2000.2222​=1.111 mol kg−1

  1. Depression in freezing point

For a non-electrolyte like glucose, van't Hoff factor i=1i = 1i=1.

ΔTf=iKfm=1×1.86×1.111≈2.067 K\Delta T_f = iK_f m = 1 \times 1.86 \times 1.111 \approx 2.067\text{ K}ΔTf​=iKf​m=1×1.86×1.111≈2.067 K

  1. Freezing point of the solution

Tf=273.15−2.067=271.083 KT_f = 273.15 - 2.067 = 271.083\text{ K}Tf​=273.15−2.067=271.083 K

  1. Nearest integer

Tf≈271 KT_f \approx 271\text{ K}Tf​≈271 K

Therefore, the freezing point of the solution is 271 K271\text{ K}271 K.

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