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CBSE Class 12 Chemistry 2016 Delhi Set 1 Solved Paper

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Question : 26 of 26
Marks: +1, -0
(a) Calculate the freezing point of solution when 1.9 g1.9\,\text{g} of MgCl2(M=95 g mol−1)\mathrm{MgCl}_2 ( M=95\,\text{g}\,\text{mol}^{-1} ) was dissolved in 50 g50\,\text{g} of water, assuming MgCl2\mathrm{MgCl}_2 undergoes complete ionization. (Kf.( K_f. for water =1.86 K kg=1.86\,\text{K}\,\text{kg} mol−1\text{mol}^{-1} )
(b) (i) Out of 1 M1\,\text{M} glucose and 2 M2\,\text{M} glucose, which one has a higher boiling point and why?
(ii) What happens when the external pressure applied becomes more than the osmotic pressure of solution?
OR
(a) When 2.56 g2.56\,\text{g} of sulphur was dissolved in 100 g100\,\text{g} of CS2\mathrm{CS}_2, the freezing point lowered by 0.383 K0.383\,\text{K}. Calculate the formula of sulphur (SX)( S_{X} ). (Kf.( K_f. for CS2=3.83 K kg mol−1\mathrm{CS}_2=3.83\,\text{K}\,\text{kg}\,\text{mol}^{-1}, Atomic mass of Sulphur =32 g mol−1)=32\,\text{g}\,\text{mol}^{-1} )
(b) Blood cells are isotonic with 0.9%0.9\% sodium chloride solution. What happens if we place blood cells in a solution containing;
(i) 1.2%1.2\% sodium chloride solution ?
(ii) 0.4%0.4\% sodium chloride solution ?.
Solution:  
(a)   WB(  solution  )=1.9 g ws\; W_B( \; \text{solution} \; ) = 1.9\,\text{g}\,ws
  W(H2O)=50 g\; W ( \mathrm{H_2O} ) = 50\,\text{g}
  M(MgCl2)=95 g/mol\; M ( \mathrm{MgCl}_2 ) = 95\,\text{g}/\text{mol}
  i=3(  for  MgCl2.  )\; i=3 ( \; \text{for} \; \mathrm{MgCl}_2. \; \text{)}
  m=WB  in  kg  Molar mass  ×  mass of solvent  \; m = \frac{ W_B \; \text{in} \; \text{kg} }{ \; \text{Molar mass} \; \times \; \text{mass of solvent} \; }
  ΔTf=iKfm\; \Delta T_f = iK_f m
  ΔTf=3×1.86×1.995×100050(kg)\; \Delta T_f = 3 \times 1.86 \times \frac{1.9}{95} \times \frac{1000}{50} ( \text{kg} )
  ΔTf=2.232 K\; \Delta T_f = 2.232 \, \text{K}
  ΔTf=ΔTf∘−Tf\; \Delta T_f = \Delta T_f^{\circ} - T_f
  Also,  ΔTf=ΔTf∘−Tf\; \text{Also,} \; \Delta T_f = \Delta T_f^{\circ} - T_f
  Tf=Tf∘−ΔTf=273.15−2.232\; T_f = T_f^{\circ} - \Delta T_f = 273.15 - 2.232
  =270.918 K\; = 270.918 \, \text{K}
(b) (i) 2 M2\,\text{M} glucose, because more will be the concentration molality, more will be the elevation in boiling point.
(ii) Reverse osmosis takes place.
OR
(a) ΔTf  =iKfm\Delta T_f \; = i K_f m
ΔTf  =Kf×WBWA(kg)×1WA(kg)\Delta T_f \; = K_f \times \frac{ W_B }{ W_A ( \text{kg} ) } \times \frac{1}{ W_A } ( \text{kg} )
MB  =  KfΔTf×WA(kg)M_B \; = \; \frac{ K_f }{ \Delta T_f \times W_A ( \text{kg} ) }
  =  3.83×2.56×10000.383×100(kg)\; = \; \frac{ 3.83 \times 2.56 \times 1000 }{ 0.383 \times 100 ( \text{kg} ) }
  =256 g/mol\; = 256\,\text{g}/\text{mol}
MB  =n×  Atomic massM_B \; = n \times \; \text{Atomic mass}
n  =  MBn \; = \; \frac{ \text{M} }{ B }
  =  25632=8\; = \; \frac{256}{32} = 8
∴S8\therefore \mathrm{S}_8
(b) (i) Water moves out from blood cell, hence will shrink.
(ii) Water will enter into blood cell, hence will swell.
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