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CBSE Class 12 Chemistry 2017 Outside Delhi Set 1 Solved Paper

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Question : 25 of 26
Marks: +1, -0
(a) A 10%10\% solution (by mass) of sucrose in water has a freezing point of 269.15 K269.15\ \text{K}. Calculate the freezing point of 10%10\% glucose in water if the freezing point of pure water is 273.15 K273.15\ \text{K}.
Given :
(Molar mass of sucrose =342 g mol−1=342\ \text{g}\ \text{mol}^{-1} )
(( Molar mass of glucose =180 g mol−1=180\ \text{g}\ \text{mol}^{-1} )
(b) Define the following terms :
(i) Molality ( mm )
(ii) Abnormal molar mass
OR
(a) 30 g30\ \text{g} of urea (M=60 g mol−1)(\text{M}=60\ \text{g}\ \text{mol}^{-1}) is dissolved in 846 g846\ \text{g} of water. Calculate the vapour pressure of water for this solution if vapour pressure of pure water at 298 K298\ \text{K} is 23.8 mm Hg23.8\ \text{mm}\ \text{Hg}.
(b) Write two differences between ideal solutions and non-ideal solutions.
Solution:  
(a)   ΔTf=Kfm\;\Delta T_f = K_f m
     Here,   m=w2×1000M2×M1\;\; \text{ Here, }\; m=w_2 \times \frac{1000}{M_2} \times M_1
  273.15−269.15    =Kf×10\;273.15-269.15\;\; = K_f \times 10 ×1000342×90\times \frac{1000}{342} \times 90
  Kf=12.3 K kg/mol\; K_f=12.3\ \text{K}\ \text{kg}/\text{mol}
  ΔTf=Kfm\;\Delta T_f = K_f m
    =12.3×10×1000180×90\;\; =12.3 \times 10 \times \frac{1000}{180} \times 90
    =7.6 K\;\; =7.6\ \text{K}
  Tf=273.15−7.6=265.55 K\; T_f=273.15-7.6=265.55\ \text{K}
(or any other correct method)
(b) (i) Number of moles of solute dissolved in per kilogram of the solvent.
(ii) Abnormal molar mass : If the molar mass calculated by using any of the colligative properties to be different than theoretically expected molar mass.
OR
(a) (i) (PA0−PA)PA0  \frac{(P_A^0 - P_A)}{P_A^0}\; =(wB×MA)(MB×wA)= \frac{(w_B \times M_A)}{(M_B \times w_A)}
  23.8−PA23.8  =(30×18)60×846\;\frac{23.8-P_A}{23.8}\; = \frac{(30 \times 18)}{60} \times 846
  23.8−PA=23.8×[(30×18)60×846]\;23.8 - P_A = 23.8 \times \left[ \frac{(30 \times 18)}{60} \times 846 \right]
  23.8−PA=0.2532\;23.8 - P_A = 0.2532
(b)
 Ideal solution  Non-ideal solution
 (a) It obeys Raoult's law over the entire range of concentration.  (a) Does not obeys Raoult's law over the entire range of concentration.
 (b) Δm i xH=0\Delta_{\text{m i x}} H = 0  (b) Δm i xH\Delta_{\text{m i x}} H is not equal to 0 .
 (c) Δm i xV=0\Delta_{\text{m i x}} V = 0  (c) Δm i xV\Delta_{\text{m i x}} V is not equal to 0 .
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