there is a key piece of information that we are missing.
we need the following:
Kb of water= 0.512
the change in boiling point (ΔTb) can be calculated using the following formula:
ΔTb= Kb x m
we already have Kb, but we need to determine the molality (m).
[tex]molality (m)= \frac{moles glucose}{Kg H_2O} [/tex]
1) let's convert the grams of glucose to moles using the molar mass of it. The molecule formula of glucose is C₆H₁₂O₆.
molar mass C₆H₁₂O₆= (6 x 12.0) + (12 x 1.01) + (6 x 16.0)= 180 g/mol
[tex]14.7g ( \frac{1 mol}{180 grams} )= 0.0817 moles[/tex]
2) let's determine the Kilograms of water.
info:
density of water= 1.0 g/ mL or 1 grams = 1 mL
1000 grams= 1 kilogram
[tex]150.0 mL = 150.0 grams[/tex]
[tex]150.0grams ( \frac{1 kg}{1000 grams} )= 0.1500 Kg[/tex]
3) let's plug in the values to solve for molality
[tex]molality= \frac{0.0817 moles}{0.1500 Kg} = 0.545 m[/tex]
finally, we can solve for change in boiling point.
ΔTb= Kb x m
ΔTb= (0.512) (0.545m)= 0.279°C
