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CBSE Class 12 Chemistry 2018 Solved Paper

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Question : 6 of 26
Marks: +1, -0
Calculate the freezing point of a solution containing 60 g60\ \mathrm{g} of glucose (Molar mass =180 g mol−1=180\ \mathrm{g}\ \mathrm{mol}^{-1} ) in 250 g250\ \mathrm{g} of water. ( Kf(\ K_{f} of water =1.86 K kg mol−1)    2=1.86\ \mathrm{K}\ \mathrm{kg}\ \mathrm{mol}^{-1})\;\;2
Solution:  
ΔTf  =Kfm\Delta T_f\; = K_f m
  =Kf×w2×1000M2×w1\; = K_f \times \frac{w_2 \times 1000}{M_2 \times w_1}
  =1.86×60×1000180×250\; = \frac{1.86 \times 60 \times 1000}{180 \times 250}
  =2.48 K\; = 2.48\ \mathrm{K}
ΔTf  =Tf∘−Tf\Delta T_f\; = T_f^{\circ} - T_f
2.48  =273.15−Tf2.48\; = 273.15 - T_f
Tf  =273.15−2.48=270.67 KT_f\; = 273.15 - 2.48 = 270.67\ \mathrm{K}
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