18 g of glucose is dissolved in 1 kg of water in a beaker. At what tem

1.	18 g of glucose is dissolved in 1 kg of water in a beaker. At what temperature will water boil at 1.013 bar? (Kb for water is 0.52 K kg mol-1 )
Answer» Number of moles of glucose = 18/180 = 0.1 mol. Mass of solvent = 1 kg. Morality of glucose , m = \(\frac{n_8}{W_A}\) = \(\frac{0.1\,mol}{1\,kg}\) = 0.1 mol/Kg Elevation of boiling point ΔT_b = K_b × m = 0.52 K kg/mol × 0.1 mol/kg = 0.052 K Since water boils at 373.15 K at 1.013 bar pressure, the boiling point of solution will be 373.15 K + 0.052 K = 373.202 K

18 g of glucose is dissolved in 1 kg of water in a beaker. At what temperature will water boil at 1.013 bar? (Kb for water is 0.52 K kg mol-1 )

Answer»

Number of moles of glucose = 18/180 = 0.1 mol.

Mass of solvent = 1 kg.

Morality of glucose , m = \(\frac{n_8}{W_A}\)

= \(\frac{0.1\,mol}{1\,kg}\)

= 0.1 mol/Kg

Elevation of boiling point ΔT_b = K_b × m

= 0.52 K kg/mol × 0.1 mol/kg = 0.052 K

Since water boils at 373.15 K at 1.013 bar pressure, the boiling point of solution will be

373.15 K + 0.052 K = 373.202 K

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18 g of glucose is dissolved in 1 kg of water in a beaker. At what temperature will water boil at 1.013 bar? (Kb for water is 0.52 K kg mol-1 )

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