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NCERT Class XII Chemistry
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Question : 51 of 53
Marks: +1, -0
Boiling point of water at 750 mm Hg is 99.63°C. How much sucrose is to be added to 500 g of water such that it boils at 100°C ? Molal elevation constant for water is 0.52 K kg mol−1^{-1}.
Solution:  
Elevation in boiling point (ΔTb)=100−99.63=0.37∘ C(\Delta T_{b})=100-99.63=0.37^{\circ}\,\mathrm{C}
Mass of solvent (water), w1=500 gw_{1}=500\,\mathrm{g} Mass of solute w2=?w_{2}=?
Molar mass of solvent, M1=18 g mol−1M_{1}=18\,\mathrm{g\,mol}^{-1}
Molar mass of solute, C12H22O11=342 g mol−1\mathrm{C}_{12}\mathrm{H}_{22}\mathrm{O}_{11}=342\,\mathrm{g\,mol}^{-1}
Applying the formula, ΔTb=Kb×w2×1000M2×w1\Delta T_{b}=K_{b}\times\frac{w_{2}\times1000}{M_{2}\times w_{1}}
or, w2=M2×w1×ΔTb1000×Kbw_{2}=\frac{M_{2}\times w_{1}\times\Delta T_{b}}{1000\times K_{b}}
=342×500×0.371000×0.52=122 g=\frac{342\times500\times0.37}{1000\times0.52}=122\,\mathrm{g}
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