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CBSE Class 12 Chemistry 2016 Outside Delhi Set 1 Solved Paper

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Question : 15 of 26
Marks: +1, -0
Calculate the boiling point of solution when 4 g4\,\text{g} of MgSO4  (M=120 g mol−1)\mathrm{MgSO}_4\;(M=120\,\text{g}\,\text{mol}^{-1}) was dissolved in 100 g\text{g} of water, assuming MgSO4\mathrm{MgSO}_4 undergoes complete ionization.
(Kb(K_b for water =0.52 K kg mol−1)=0.52\,\text{K}\,\text{kg}\,\text{mol}^{-1})
Solution:  
ΔTb  =iKb⋅m\Delta T_b\;=i K_b \cdot m
i  =2i\;=2
  =i×Kb×W2×1000M×W1\;=i \times K_b \times \frac{W_2 \times 1000}{M \times W_1}
  =2×0.52 K kg mol−1\;=2 \times 0.52\,\text{K}\,\text{kg}\,\text{mol}^{-1} ×4 g×1000 g/kg120 g/mol×100 g\times \frac{4\,\text{g} \times 1000\,\text{g/kg}}{120\,\text{g/mol} \times 100\,\text{g}}
  =2×0.523\;=\frac{2 \times 0.52}{3}
  =0.346 K\;=0.346\,\text{K}
Boiling point of water  =373.15 K373 K\text{Boiling point of water}\;=\frac{373.15\,\text{K}}{373\,\text{K}}
Tb  =Tb0+ΔTbT_b\;= T_b^0 + \Delta T_b
  =373.15 K+0.346 K  or  \;=373.15\,\text{K} + 0.346\,\text{K}\;\text{or}\; 373 K+0.346 K373\,\text{K} + 0.346\,\text{K}
  =373.496 K  or  373.346 K\;=373.496\,\text{K}\;\text{or}\;373.346\,\text{K}
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