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CBSE Class 12 Chemistry 2013 Solved Paper

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Question : 9 of 30
Marks: +1, -0
18 g18\text{ g} of glucose, C6H12O6(.\mathrm{C}_6\mathrm{H}_{12}\mathrm{O}_6 (. Molar Mass =180 g mol−1)=180\,\mathrm{g}\,\mathrm{mol}^{-1}) is dissolved in 1 kg1\text{ kg} of water in a sauce pan. At what temperature will this solution boil?
(Kb.(K_b. for water =0.52 K kg mol−1=0.52\,\mathrm{K}\,\mathrm{kg}\,\mathrm{mol}^{-1}, boiling point of pure water =373.15 K=373.15\,\mathrm{K} )
Solution:  
ΔT=Kb×WB×1000mB×WA\Delta T = \frac{K_b \times W_B \times 1000}{m_B \times W_A}
ΔT=0.52×18×1000180×1000=0.052\Delta T = \frac{0.52 \times 18 \times 1000}{180 \times 1000} = 0.052
T−Tb=0.052T - T_b = 0.052
T−373=0.052T - 373 = 0.052
T=373+0.052T = 373 + 0.052
T=373.052 KT = 373.052\,\mathrm{K}
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