Practice question
Question
Calculate the boiling point elevation of a solution containing 7.2 g of glucose ( C₆H₁₂O₆ ) in 300 g of water. ( K_b = 0.52 K kg/mol, Molar mass of glucose = 180 g/mol )
Explanation
Given:
Calculate the boiling point elevation of a solution containing 7.2 g of glucose ( C₆H₁₂O₆ ) in 300 g of water. ( K_b = 0.52 K kg/mol, Molar mass of glucose = 180 g/mol )
Formula:
Moles of glucose = 7.2/180 = 0.04 mol.
Substitution & Calculation:
Molality = 0.04/0.3 = 0.133 mol/kg . Δ T_b = 0.52 × 0.133 = 0.069 K .
Final Result:
The computed value matches expected outcome and confirms correct choice as per latest NCERT 2026-27.
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