58.5 g of NaCl and 180 g of glucose were separately dissolved in 1000 mL of water. Identify the correct statement regarding the elevation of boiling point (bp) of the resulting solution.

Elevation in boiling point, ∆Tb=i×Kb×mMolality of NaCl solution =nw×1000                                       =58.558.5wH2o×1000=1000wH2oMolality of C6H12O6solution =180180×1000wH2O=1000wH2OBoth solutions have same molality but values of i i.e. van't Hoff factor for NaCl and glucose are 2 and 1 respectively.Hence, NaCl will show higher elevation in boiling point.